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Questions and Answers

Here is a history of questions and answers processed by "Ask the Physicist!". If you like my answer, please consider making a donation to help support this service.

If there is a link to a previously answered question, be patient. Since the files containing the older answers are rather large, it takes some time (maybe as much as 15 seconds or so) to find the appropriate bookmark.


QUESTION:
According to newtons 3rd law of motion, a rock sitting on the ground pushes against the groung, and the ground pushed back against the rock with force. explain why this force doesnt cause the rock to accelerate upward from the ground according to newtons 2nd law.

ANSWER:
You need to look at all the forces on the rock. Let us say that it has a weight of 5 lb. Then the earth pulls down on the rock with a force of 5 lb (that is what weight means). But Newton's first law says that if the rock is in equilibrium (which it clearly is), the net force on it must be zero. Since the only other thing which can possibly exert a force on the rock is the ground, it must exert a force up on the rock which is equal in magnitude to 5 lb making the net force zero. Nothing needs to be said about the third law. If you want to invoke the third law, then the fact that the ground pushes up on the rock with a force of 5 lb, the rock must push down on the ground with a force of 5 lb. You could also say that if the earth pulls down on the rock with a force of 5 lb, the rock must pull up on the earth with a force of 5 lb.


QUESTION:
My nephew asked me why snow is white.

ANSWER:
There is a good explanation at HowStuffWorks.com.


QUESTION:
My son wanted to do a science project on time dilation, we read that it can only be proved with thought expts., like twin paradox. I also read that some expt. Can be made with two parallel mirrors[when light is passed through the mirror it reflects and bounces making a v_shape pattern]., which proves speed of light differs in different frames. Is this a proper expt. To prove time dilation or is not a proper proof?, pls let me know, my son is 8 yrs old he is very much interested onthis topic.

ANSWER:
What does "proved" mean to you? I think you want, for an 8-year old, to make it plausible and demonstrable, not proven. Proven, I guess, would include a derivation of the Lorentz transformation from which time dilation follows naturally, and that is surely far beyond your son. The first thing you need to do is to get him to appreciate the one thing you must accept in order to understand relativity, that the speed of light is the same no matter who measures it. So, if you shine a light forward from your rapidly moving spaceship, you will measure its speed to be 186,000 miles/second and, even if your speed is 93,000 miles/second, somebody you are going by will still measure the speed of the light to be 186,000! You will find this discussed on my FAQ page and will probably need to paraphrase those discussions for your son. Now, how can you make time dilation plausible, given the constancy of the speed of light? That is where the light clock you allude to comes in. But this experiment does not "prove[s] speed of light differs in different frames", quite the contrary: it assumes that the speed of light does not differ and shows that it follows that moving clocks run slower than stationary clocks (time dilation). But, is it all academic, we can only "observe" it for "thought experiments" (traditionally called by the German name gedanken)? No, the theory of special relativity is one of the best verified in all of physics, never has there been any indication that any of it is wrong. But to easily see time dilation in action requires very high speeds. For example, consider a pi meson, an elementary particle which has a half life of 18 nanoseconds (10-9 s). Suppose we make one in the lab with a speed of 80% the speed of light. Since the speed of light is 3x108 m/s, we expect it to go (3x108)(18x10-9)=5.4 m. But when we actually observe the pi meson, it goes 9 m. Why? Because this particle is like a clock that ticks once and dies; that clock is moving very fast and therefore ticks much more slowly than if it were just sitting here, so it can go a lot farther than we expect it to. Another good example is the satellites used in GPS systems. Their speeds are quite small compared to light speed, but it is extremely important to measure time very accurately for the systems to be accurate; if corrections for time dilation were not applied, you would never be able to have GPS with near the resolution we all take for granted.


QUESTION:
A steel railing on my outside deck connects to the house near our bedroom. When it is very cold outside the railing vibrates - a very low harmonic hum, almost like a train in the far distance. The sound keeps us awake and can cause a headache. I have solved it by strapping a heavy piece of carpet over the railing which diminishes the vibration and noise. This only occurs in the cold with a wind. Never the rest of the year. Why?

ANSWER:
Wind can cause things to vibrate. If the object is free to vibrate and the wind causes vibrations of one or more of the resonant frequencies of the object, the result can be quite dramatic. The classic example, of course, is the collapse of the Tacoma Narrows bridge. Since most materials expand when heated, it would appear that something is contracting which allows the railing to vibrate when it is cold. I do not know how it is attached to whatever it is attached to, but I would check to see if the attachments could be tightened up when it is cold. That might clamp the vibrations. If not, your carpet (which is called damping) may be your only option.


QUESTION:
I've been wondering if I had a telescope fixed on a star that has gone super nova, but the light of the explosion had not yet reached earth, if I mounted said telescope to a rocket traveling close to light speed and flew towards the star would I see the evolution of the star to it's final fiery demise being accelerated due to: a. The fact that I am traveling towards the light source b. The effects of time dilation Also, would it begin to glow with a blueish tint due to the doppler effect?

ANSWER:
This is another question which asks "how fast do clocks appear to run?" You will indeed see the supernova you are moving toward as happening sooner and faster than if you watched from earth. It also will be blue-shifted because of your motion. To understand this better, consider the return trip of an earlier answer.


QUESTION:
How electrons & protons combine to form neutrons in a neutron star?

QUESTION:
What should happen when electrons are squeezed into the nucleus?

ANSWER:
What happens is the inverse of beta decay, a proton and electron go to a neutron and a neutrino: p+e>n+
νe.


QUESTION:
for almost five years since we were taught, we've known that current is a scalar quantity having direction that does not obey vector laws, but, obeys Kirchoff's laws. However, I'm recently studying electrodynamics from Griffiths and it says that current is vector quantity. Its very confusing. If current is vector, it has magnitude and direction that obeys vector laws and also directions with respect to Kirchoff's laws. My feeling is that it is vector at microscopic scale and scalar at macroscopic scale. But, how come?

ANSWER:
Current density J is a vector. The current through some area A is defined as A
JdA over the area. So, current can be either positive or negative depending on the sign of the dot product. It is not a vector, but its sign matters, particularly when applying Kirchoff's laws as you have found.


QUESTION:
If a rocket is in space and you add thrust to the rocket to make it move forward for 10 seconds, and then 2 minutes later you add the same amount of thrust to move the rocket forward again. Since you are already going X MPH would the rocket increase speed based on the additional thrust or would it stay the same speed since you are already going X MPH? Being that there is no friction (very little) in space, would each thrust forward add to the forward motion and increase the speed? Also meaning that if you continued to add thrust without stopping, even in small amounts, you would continuously go faster and faster until you ran out of thrust? I undertsand that the faster an object goes, the more mass it has, but I do not understand why that would need additional thrust if there is no friction in space.

ANSWER:
All this turning on and turning off "thrust" just complicates things. Let's focus on a simpler situation where we simply apply a constant force to the rocket continuously. In classical physics, this would just result in the rocket going faster and faster without bound at a constant rate. However, when the speed becomes comparable to the speed of light, we are constrained to not exceed the speed of light and acceleration ceases to be constant even though the force is constant. For a detailed discussion of this problem, see an earlier answer.


QUESTION:
If Earth is moving through Space at an incredible rate, why do relatively tiny changes in speed effect humans so strongly on earth? If Earth is moving through the Universe, within the galaxy, within whatever larger system, at astronomical speeds, how does something like a roller coaster or a plane ride - which changes the speed from not moving on Earth so incredibly fractionally compared to the overall speed we are travelling through the universe - have such a great effect on the Human Body.

ANSWER:
The way that the earth moves subjects you to very small accelerations and acceleration is what you feel.

FOLLOWUP QUESTION:
So we know for a fact that the earth isnt accelerating or decelerating through space any fast than a fast car pulling away from a stop light? In fact, significantly less? That seems rather improbable to me. How in the world could it be that we can easily achieve acceleration noticeable enough to feel on the ultra-small scale of earth - such as in falling off a cliff, or skydyving - while the enormous earth in an enormous solar system, in an enormous galaxy, in an enormous universe that's expanding at an unbelievable rate somehow reflects an acceleration less than this - unnoticeable to us. To put it another way, if the universe is expanding, how could it possible be that the de/acceleration of the universe is less than the meager capabilities of a motorbike.

ANSWER:
You confuse speed with acceleration. Just because something is going very fast, it is not necessarily accelerating. What matters is the rate at which a velocity vector is changing. The greatest acceleration due to the external sources you list is due to the earth's rotation. The speed at the equator is about 465 m/s which is about 1040 mph, pretty fast compared to your motor bike. But, what is the acceleration related to this motion? It is about 0.034 m/s2 (from v2/R), about 300 times smaller than the acceleration due to gravity. Another example is the acceleration due to earth's orbit around the sun. The speed here is much bigger, about 30,000 m/s, but the acceleration is about 0.006 m/s2.


QUESTION:
I would be grateful if you might help me with a difficult problem in mechanics that I have tried to solve for a long time without success. The problem is a part of a bigger problem that I am trying to solve and is therefore important to me. Here is the problem: How much power P [W] is needed to force a satellite (modeled as a point particle) to move in a perfect circular orbit with radius r [m]? The mass of the satellite is m [kg] and its tangential speed v [m/s] is constant. Important conditions:

  • It is assumed that NO GRAVITY affects the satellite!

  • The mass loss due to fuel consumption is neglected.

  • The power is assumed to act under ideal conditions, i.e. there is no energy loss due to friction or the choice of engine.

  • It is also assumed that v << c, i.e. relativistic effects are neglected.

I have received many answers saying that P = 0 as a consequence of the definition of mechanical work. But this cannot be true since no gravity affects the satellite. Energy is needed to change a linear path in space and a circular orbit is one such motion. In absence of gravity no particle with mass will loop in a circle without any energy.

ANSWER:
What you have been told before is correct. No work is done by the force causing the satellite to move in orbit. Since no work is done, we can ask whether the energy is conserved. Since there is no gravity, the only energy to think about is the kinetic energy of the satellite and since the speed is the same always, so is the kinetic energy. Or, look at it this way: suppose we are in otherwise empty space but that there is a fixed point to which we can tie a rope, the other end of which we tie to the satellite which we give a shove so that it moves in a circle around the center. Does the string need to supply any energy? Do you need to connect a machine of some sort to the rope? Your basic premise, "Energy is needed to change a linear path in space", is simply wrong; a force is needed, but not energy. This is not a "difficult problem".


QUESTION:
Have there been any theories presented by rational and peer-accepted scientists, trying to pair the combining force of Gravity with the separating force of Dark Energy into a single equation or system of equations?

ANSWER:
This already exists, has existed ever since the origin of general relativity. At the time Einstein developed general relativity to explain gravity, it was believed that the universe was static. He therefore included a term in the general relativistic equations called the cosmological constant. Later, when Hubble found that the universe was expanding, he removed the constant calling it his "greatest blunder". If now it is reinserted, dark energy is simulated. However, this is far from a complete theory or explanation, seems to be pretty empirical.


QUESTION:
A friend claims that the inside of an LP record (or CD for that matter) spins faster than the outside. I say this is impossible because if it were true, the record/CD would tear itself apart. Who is right?

ANSWER:
LPs and CDs spin differently and therefore must be discussed separately. First we need to clarify what "spin faster" means. Since a disk is a rigid object, each point has the same angular velocity, revolutions per minute (rpm) for example. Each point on any disk rotating at 100 rpm goes around 100 times per minute. However, if you look at the speed of any point on the disk, the farther out you go the faster the point goes; a point 4 inches from the axis will move with a speed twice as large as a point 2 inches from the axis. An LP is sometimes referred to as a "33" because it always spins with an angular velocity 33 1/3 rpm. The groove moves past the needle slower and slower as the needle moves in during the playing. So, if you were to microscopically look at what the groove looked like for some note, that would look different depending where on the record that note was recorded. CDs, on the other hand, are designed so that the speed with which the data on the disk pass the laser reader is always the same, so as the CD plays, the angular velocity is constantly adjusted to keep the linear velocity of the data being read the same. The CD, unlike the LP, reads from the middle out and as the laser moves out the angular velocity decreases. So, clearly neither of you is right: inside and outside spin the same, outside moves faster, CDs and LPs have different motions, and nothing tears itself apart.


QUESTION:
Kind of a simple question, I think, but just curious. If I check in at an airport with my luggage, someone said if I lay my bag down it will weigh less than verticle. Obviously the thought may not get charged for over 50lbs or whatever if I lay it down. If you're married you probably understand where I'm coming from:)

ANSWER:
The pressure the bag exerts will be smaller, but the weight is unchanged. The scale they use at check in measures weight, not pressure. No comment on the marriage statement.


QUESTION:
First, take a large round, flat-bottomed container and pour in 3 inches or more of water. Sprinkle in about a quarter to a half teaspoon of sand. First stir the water at random and notice how the sand distributes itself in a random manner. Then, swirl the water around in the container with a spoon or other utensil in a continuous clock-wise or counter clock-wise motion for several seconds. Remove the stirrer. Why do the particles of sand aggregate in the center of the vortex? Since they are heavier than water (they sink), why aren't they spun to the sides of the container due to the negative centripetal force? Is this behavior similar for air and, for example, saw dust in industrial dust collectors?

ANSWER:
Seems it is not a trivial question. There is a ton of stuff here, all of which I have not waded through.

FOLLOWUP:
I followed up on the reference to "Einstein's Tea Leaves" and found a very good article.


QUESTION:
At what velocity and angle would one have to launch a paintball from the surface of the moon for it to achieve and maintain an orbit? Paintballs weigh approx 4 grams and are approx .685" in diam.

ANSWER:
First of all, the mass and size of the paintball is irrelevant. The orbital speed for any particular orbit is the same for any mass (small compared to the mass of the moon). And, you cannot launch from the surface if you want it to not eventually hit the surface again, you must launch from above the surface. But let's say that we launch horizontally a few feet above the surface (and neglect any mountains it might hit). The minimum launch speed would be about 1680 m/s
≈3760 mph.


QUESTION:
While using an elliptical machine for exercise, as one increases the resistance from the machine for a period of time, does one actually exert more force with his legs? Or is the maximum force a person can exert just the force from his body weight (gravity) which doesn't change as he exercises? Can one press harder with one's legs if he doesn't have something to brace against to increase that force?

ANSWER:
I do not know the details of an elliptical machine, have never used one. But, I can tell you that you can exert a force up on your body with your legs which is greater than your weight. This is what you do when you jump. The fact is that if you do so, your torso accelerates up. For example, if you are in a squat and stand up quickly your legs are exerting a force greater than your weight; the quicker  you stand up, the bigger the force. I do not know what it means to "increase the resistance" of the machine, but if it means go faster then it would be sort of like doing knee bends faster.


QUESTION:
While airing up a heavy truck tire, my friend said if I lifted the tire off the ground (relieved the load) it would air faster. This seems wrong to me. 110# is 110# (internal pressure) regardless of the external load. Who is right, and more importantly, why?

ANSWER:
Take the extreme situation as an illustrative example: when sitting on the ground the unfilled tire will be flattened down almost to the rim. This might be the case if the wheel is mounted on the truck. Now, when pumping it up, the increased air must do two things, increase the pressure and increase the volume. Increasing the volume means work is done which means it takes energy. If the tire starts out round, adding air only increases the pressure and does not do the work to increase the volume. So, I would guess that the wheel which starts out round would fill faster than one you need to expand. If the tire is already round when you start there would be no need to lift it. Nevertheless, the compressor is probably strong enough that the difference will be pretty small. Should be an easy enough thing to actually try it. Physics is an experimental science, after all.


QUESTION:
When a balloon collects a charge it gains electrons and loses the positive charge.why is it that the balloon will neve have a positive charge, everywhere I look says it will always be negative, but doesn't have an explanation.

ANSWER:
Something is charged negatively by adding electrons and positively by removing electrons. Balloons are made of rubber, an insulator. Insulators are difficult to remove electrons from because all the electrons in the atoms are tightly bound. There is no reason why electrons could not be added to an insulator, though.


QUESTION:
If I am in a rocket ship that is moving away from Earth at very near the speed of light and my friend is on another rocket ship also headed away from Earth at very near the speed of light but in the opposite direction. Do we see each other as moving away from each other at greater than the speed of light?

ANSWER:
No, because velocities do not add the way you expect them to in Galilean relativity which is v'=u+v where u and v are the speeds of the two ships and v' is the speed that one sees the other moving. This turns out to be incorrect if u and v are not small compared to the speed of light c. The relativistically correct addition formula is v'=(u+v)/[1+(uv/c2)]. So, if u=v=0.9c, v'=1.8c/(1.81)=0.994c. If you had looked at my FAQ page you would have found this discussed; the link would get you to here.


QUESTION:
I'm trying to program a robot to shoot a basketball into a basket. I've been trying to simplify doing this with only projectile motions equations (The ball weighs 11.2 oz and probably is effected by drag). The robot would be able to find the height and the distance. I was wondering if it is possible to find the initial velocity and the angle I would need, just from the height and distance?

ANSWER:
Possible, yes. Easy, no. It is a pretty complicated problem which does not have a simple formula you can use. I would try the simple model first and see how it works.

FOLLOWUP QUESTION:
So are you telling me to just use the projectile motion formulas? When I do them I end up with 3 variabled, time, initial velocity and theta

ANSWER:
I thought your concern was the air drag. So, when you do the kinematic equations, there are two:

  • One is the horizontal position which must end up at x=R where R is how far horizontally the basket is from the launch point; R=v0tcosθ where v0 is the initial velocity and t is the time of flight.

  • The other is the horizontal position which must end up at y=h where h is how far vertically the basket is above the launch point; h=v0tsinθ-½gt2 where g is the acceleration due to gravity (9.8 m/s2).

So, you have two equations with three unknowns, v0, θ, and t. The first thing to do is reduce it to one equation with two unknowns by solving the R equation for t and substituting that into the h equation. The result is h=Rtanθ-½gR2/(v0cosθ)2. Now, think about it—you cannot mathematically know both θ and v0 because you have only one equation; but you shouldn't expect to from a physical point of view either. If you change v0 you will have to change θ too to hit the target. So, there are an infinite number of right answers here, not one unique answer. Nevertheless, if you program your robot to always shoot with the same velocity, you can solve for the angle. So, you need to solve the above equation for θ and that will tell you how to aim. It is not trivial algebra, but not real hard either and I do not intend to present the general solution because it is complicated. Be aware that there will be two solutions for every situation which you can easily understand, I think. Think of a rifle shooting at a target where you have to aim a little above the target to hit it; if you also shot way up in the air, there would be some large angle where it would come back down through the target. I will leave the general solution to you; I will do one simple special case to illustrate, R=4 m, h=2 m, v0=10 m/s, and I will approximate g as 10 m/s2: 2=4sinθ/cosθ-0.8/cos2θ. Rearranging and simplifying (and using sinθ=√(1-cos2θ)), 0.4+cos2θ=2cosθ√(1-cos2θ). Now square this equation and rearrange: 5cos4θ-3.2cos2θ+0.16=0. This is a quadratic equation in cos2θ. Thus, cos2θ=(0.585 or 0.055) and cosθ=(0.765 or 0.234) and θ=(40.10 or 76.50). Of course, you could also fix θ and solve for v0. This is easier. For example, if you fixed θ at 40.10, 2=4tan40.10-80/(v0cos40.10)2 which leads to, guess what, v0=10 m/s. However, you will run into trouble here because since you are playing basketball, the ball must be coming down when it gets to the basket, and that will not always be the case for a randomly chosen set of R, h, and θ. For example, in the example I did, 40.10 would not work because the ball would be going up when it got to the basket (vy=1.2 m/s); for the other solution, 76.50, the ball would be going down (vy=-7.4 m/s) like you need to make the basket. I got these by solving for the times from the x equation and then using the velocity equation (vy=v0sinθ-gt). You have your work cut out for you!


QUESTION:
Here is a question I have long pondered: If a man is riding in the back of a pickup truck, and the truck is traveling forward at 500 feet per second, and he fires a pistol toward the rear of the truck, and the pistol fires the bullet at 500 feet per second, will the bullet leave the barrel?

ANSWER:
When you say "the pistol fires the bullet at 500 feet per second", this is the muzzle velocity, the velocity of the bullet with respect to the barrel. So, yes, the bullet will definitely leave the barrel. And if you are the man in the truck, you see the bullet go backwards with a speed of 500 fps. But, you see the rest of the world going backwards with a speed of 500 fps also. So, somebody standing on the ground will see a bullet drop straight to the ground! (This assumes we can neglect air friction.)


QUESTION:
I learned the basic premise that things moving fast can have negative effects on other things when I was in kindergarden. In elementary school I came up with a question. If light travels so fast, why doesn't it completely obliterate anything it hits, by smashing it to pieces? I had my sister ask her physics teacher. The answer he sent back to me was that light was too unfocused, and that objects with zero rest mass do not have that kind of effects on things which do have mass. (Great answer to give to a kid in elementary school.) This question remained with me as I got older, and inquiring my own physics teachers I only learned enough to become further confused about the subject. I know when something with positive rest mass, like a bullet or a baseball hits something it transfers momentum and kinetic energy. I was told that the effect light has on objects heating them etc, is an effect of kinetic energy. But I am still having trouble really understanding the concept. With learning the difference between energy, and momentum only helping slightly. If I throw a baseball with high velocity and it hits something, like a block of metal the impact of the ball will cause the block to move. If I shine a laser at the block and just keeping adding more and more power would it ever be possible to accelerate the block with the laser? (Assuming it could avoid being melted)

ANSWER:
You are interested in force so you need to ask about linear momentum (which we usually think of as p=mv, but I will modify this a little below) because, in a collision, each object experiences a force which is proportional to the change in its momentum. When a ball strikes another ball, during the time they are in contact each exerts a force on the other (which are equal and opposite, Newton's third law). The struck ball, initially at rest, rebounds forward because of this force and the striking ball slows down because of the equal/opposite force it feels. If the striking ball moves faster, it has more momentum and therefore exerts a larger average force during the collision. In fact, the force is precisely equal to the time rate of change of momentum; this is Newton's second law. Now, to your question. Light can exert a force but not because it goes fast but because it carries momentum. Even though photons have no mass, they still do have momentum. So, when light strikes something, it does exert a force on it; that force, for everyday situations, is just too small to have much of an effect because normal intensities of light just do not have all that much momentum. I will give you a couple of examples of so called light pressure. GPS satellites, to work accurately, must have their orbits calculated extremely accurately and corrections for radiation pressure from light from the sun must be applied. It has been proposed that future spaceships in the solar system could have gigantic reflecting sails and use radiation pressure as a means to be accelerated away from the sun; each photon comes in with a momentum p=E/c where E is the energy of the photon and c is the speed of light. Since the photons are reflected, their change in momentum is 2E/c which is also the momentum transferred to the spaceship.


QUESTION:
gallile said that the force of gravity is the same on everything but the force of rubbing with the air molecule that change the speed of falling,so shouldnt 2 balls made from the same materiel but one is bigger than the other falling from the same heigth,the smaller touch the ground first cz her volum is smaller than the force of rubbing should be smaller than the big ball,but that is not the fact i dnt understand?

ANSWER:
First of all, Galileo never said the force of gravity is the same on everything because the notions of force and gravity were not even conceived before he was dead! Anyhow, that is not even true; the force of gravity is greater for greater masses. What is the same is the acceleration due to gravity, which Galileo supposedly demonstrated by showing that two objects of different masses hit the ground simultaneously. This is only approximately true because there is a force due to the air resistance as the balls fall. Galileo did not take air resistance into account and balls of different masses and sizes will not hit simultaneously. If you look at my FAQ page you will find lots of discussion of problems involving air drag and you will find that the force is proportional to velocity times the area presented to the oncoming wind (approximately). If the two balls are made of the same material, their masses are in the ratio (R1/R2)3 and their cross sectional areas are in the ratio (v1R1/v2R2)2. As each ball falls, it initially is accelerating down but this acceleration gets smaller as the speed increases because of the increasing air drag. Each ball eventually falls fast enough that the air drag is equal to the weight and falls with a constant speed called the terminal velocity. So, the weight (mg) and the drag (
¼πR2v2) are equal. Taking the ratio of the equations for the two balls, (R1/R2)3=(v1R1/v2R2)2 so v1=v2(R1/R2). Suppose that #1 is the smaller ball and R2=4R1; then v1=½v2. Since the smaller ball has the smaller terminal velocity, the larger (more massive) ball will reach the ground first. So, in fact, your intuition was wrong.


QUESTION:
Given the same frontal area, mass, surface area, etc. Why is the drag coefficient less for a hemisphere than a sphere? A sphere seems more streamlined but has a larger drag coefficient.

ANSWER:
It is true that drag is approximately proportional to cross sectional area and therefore you would expect the two to have the same drag. However, the key word is approximately and computation of drag coefficients is a complicated thing and depends on aerodynamic details which can really only be determined numerically. The nature of the wake and the turbulence are crucial and often not intuitive. My favorite example is the use of tailgate nets in pickup trucks. It is obvious that opening the tailgate on a pickup truck will make the truck more streamlined, right? That is the idea behind those nets which are marketed to save gas. However, opening the tailgate considerably increases drag; the tailgate traps a "bubble of air" which moves with the truck.


QUESTION:
Often times they refer to the flipping of a coin as having an outcome that is random. If this be the case, I am wondering how then a particular side of a flipped disc had a person applied pressure to one portion of a disc and after having applied all necessary forces, i.e. wind resistance, gravity, etc., and had observed the disc up on a flat surface calculated a particular image to be heads or tails at random. I understand that it has to be either or and there is "probability" of an imagine over another. How can one be more likely than the other if you flick it the exact same way every time so that one side would never end up showing. How can probability be so disconnected from physics if even conceptually?

ANSWER:
Since flipping the coin is done by a person, it is simply impossible to "flick it the exact same way every time". Also, some of the forces, air drag in particular, have aspects which involve turbulence which is chaotic and cannot be predicted precisely nor replilcated each throw. Perhaps in a vacuum one could, by using a machine to flip the coin and having it "stick" when it hits the floor, create a situation where there would be at least a bias for heads or tails.


QUESTION:
Is there any relationship between Lennard-Jones potential and Van der Waals force ?

ANSWER:
Yes. The Van der Waals force is the long range attraction between neutral atoms or molecules. The Lennard-Jones potential represents an empirical force which has both a short range repulsion (r-12) and long range attraction (-r-6); the attraction part could be called Van der Waals.


QUESTION:
I am trying to find out how much energy was produced. When I struck. Headonatree head on The car weighted 2945 pounds travelig at 75the miles per hour. I weighed 285 pounds and was nor wearing a seat belt

ANSWER:
No energy was "produced"; the kinetic energy of your car was converted into heat, sound, and the work necessary to crumple up your car. The total kinetic energy of you and your car is
½mv2 where m is the mass (in English system units m=(285+2945)/32=101 lb s2/ft) and v is your speed (v=75x3600/5280=51 ft/s). So the total energy is ½x101x512=131,000 ft lb. That is about 0.067 kilowatt-hours; if you could take that amount of energy and spread it out over an hour, you could run a 67 watt light bulb.


QUESTION:
We learned in class that centripetal acceleration is equal to v^2/r, when velocity is constant. we also learned that acceleration is the rate of chance of velocity. Since magnitudly, velocity is not changing, wouldn't a=0? What does the value of centripetal acceleration actually mean if the magnitude of velocity is not changing? If the acceleration is 15m/s^2, but velocity is constant.. then the body is not actually accelerating at 15m/s^2. So basically what does the magnitude of centripetal acceleration actually tell you? What does it mean?

ANSWER:
Velocity is a vector, that is it has both a magnitude (usually called speed) and a direction. Acceleration is the time rate of change of velocity and velocity can change by either changing its magnitude or its direction (or both). An object in uniform circular motion is accelerating because the direction of the velocity, which has constant speed, is changing. Acceleration is also a vector and its direction is toward the center of the circle for uniform circular motion. The problem is that in everyday life we usually think of acceleration as something speeding up but that is not the general meaning in physics.


QUESTION:
I have learned that every subatomic particle behaves in a wave like particle that can described using the wave equations, however, what exactly does it mean when they say waves? I know photons travel in waves, but how do protons, electrons, and other particles exhibit wave behavior.

ANSWER:
It is one of the tenets of quantum mechanics that any particle is also a wave. Look for a particle, you will find particle-like properties, look for a wave, you will find wave-like properties. This was first proposed by Louis de Broglie in 1924 and he won the Nobel prize for his hypothesis in 1929. It has subsequently been verified innumerable times in experiments. For one example of diffraction of protons by a sphere, see an earlier answer.


QUESTION:
I am a Stuntman for TV and film and we have created a device called a ratchet. It is a pneumatic piston that very quickly pulls a line (rope, cable etc.) to "yank" someone or something a certain distance. (explosions, etc...) So, if I had a 200 lb man at the end of a line, what dynamic force is generated on the line at 6 ft per second. (no ground dragging friction, it generally would pull you up in a 45 degree angle with the aid of rigging and pulleys). I have heard that this can multiply by 4 or 5 creating an dynamic load of 800 or 1000lb on the line for that instant. is that accurate, more, less? or is there a formula. please, if possible, answer in layman terms.

ANSWER:
What matters is not the speed, but how quickly you get there. Once you get to constant speed, the tension in the line will always be less than or equal to the weight. It is easiest to understand if we keep it simple and lift vertically. Just to hold the 200 lb man vertically obviously means the tension will be 200 lb. But, suppose we want to accelerate him to 6 ft/s in 0.1 s; his acceleration is 60 ft/s2. Newton's second law says that there must be a net force upward which is equal to his mass times his acceleration. This is a little tricky using English units (ft, lb, s), but without any details, just accept that his mass is 200/32. Therefore the net force needed is (200/32)x60=375 lb. However, there is already a 200 lb force down (his weight), so the tension in the line would have to be 375+200=575 lb, almost three times his weight. Once you stop accelerating him, it makes no difference what his speed is, the tension in the line will be 200 lb. Pulling horizontally (ignoring friction) the tension would only be 375 lb, so pulling at 450 would be somewhere between. You could replicate this calculation for any acceleration you wanted; 0.1 s is pretty fast, but the man could take more since 60 m/s2 is only about 2g's of acceleration.


QUESTION:
What is the minimum mass that an object can be in order to bend the path of a photon? Does the bending of the path result in a decrease in the speed of the photon?

ANSWER:
Any mass creates a gravitational field and any gravitational field will result in a photon being deflected. However, for the deflection to be big enough to observe the mass has to be really huge. The speed of a deflected photon is unchanged.


QUESTION:
I recently helped my 11 year old son design and test a science project. The project hoped to measue a quantifiable difference between a "corked" baseball bat and a non-corked bat. I can't help but thnk I did not consider all the necesarry factors. We have only limited resources so we tried to make it as simple as possible. We built a gantry and mounted it upon a rectagualr base. From the gantry the bat swings from a vertical position about 150 degrees above the contact point of bat on ball, hitting the ball which has been placed on a tee. It is hit straight out from the contact point onto a long sheet of waxed paper (which records the spot where the ball hit), The spots where the balls hit are measured and recorded. In an attempt to create an experiment that uses a constant bat speed, we decided to use gravity as the "force generator" since it would be the same every time. One of the "beliefs" about the benefit of corking a bat is that the reduced weight of the bat allows the bat to be swung faster/harder (at least as I understand it). I don't know if that is in fact the case but, (here's my question): Did I err by making gravity the "constant" in my experiment? Galileo showed that two objects of different weight arrive at the ground at the same time. (haha-I think) So the two bats we are testing will fall through their arc at the same rate of speed, won't they? One wil not fall faster because it is lighter will it? I can't see how this differs from a baseball player swinging a bat horizontaly and having it move faster becasue it is lighter.

ANSWER:
The main problem here is that you are not replicating Galileo's dropping two balls experiment. Your situation is one of rotational motion, not simple falling, and the effect of gravity on the two bats is going to be different. Since your son is only 11, I will try to keep it nonmathematical. The time it takes the bat to drop (and the speed it will have at the bottom) is not determined by what the mass is but it does depend on how that mass is distributed. Think about the center of mass* of each bat; since the corked bat is lighter at the extreme end of the barrel, its center of mass is closer to the knob than the uncorked bat. So, as a first approximation, think of each as a simple pendulum, a tiny point mass on a string, where the point mass is at the center of mass and the string is thought of to run between the hinge and the center of mass. A simple pendulum with a short string has a shorter period than one with a longer string, so the corked bat should actually be moving faster at the bottom than the uncorked bat because its center of mass is closer to the hinge.  And, if you had a video camera and could examine its movie frame by frame, you could determine the speed of each bat at the time it hit the ball.

Again, this is a rough approximation because neither is really a simple pendulum as I have used to try to make it 11-year old friendly (you can easily make simple pendula and show him that the shorter gets to the bottom quicker). In real life, these are called physical pendula and you have to worry about how the mass is distributed in detail (moment of inertia if you have ever studied physics), not just where the center of mass is, but I think my explanation should give a good qualitative overview.

So, the answer to your final question is that one bat does fall faster but not because it is lighter but because its mass is differently distributed.

*You can easily find where the center of mass of each bat is by finding where each balances.


QUESTION:
My Son & I were are having a discussion about how far a golfball would go on the moon vs. the Earth, if hit @ a 45 degree angle. The ball, club, strike force and any other possible variable would be identical with the exception of gravity and atmospheric resistance. The accepted story is the golfball would travel the equivalent of the difference between the Earth and Moons' gravitational pull as the multiplier or divisor. I contend the lack of atmospheric drag on the moon would cause the golfball to to go measurably farther. What say you?

ANSWER:
I say that the lack of air on the moon gives the moon ball more advantage than gravity alone. You are right.


QUESTION:
Suppose a spherical ball is tied wiht a string , is moved in a horizontally circular path with other end as centre . Now if we apply more force to the end of string at centre to move the ball faster , then what force is responsible for speeding up the ball? However all the forces are perpendicular to the tangential velocity.

ANSWER:
If you apply more force to the end of the string, the ball will be pulled into a path of smaller radius. So, the velocity of the ball will have a component in the direction of the force and so work is done to speed up the ball.


QUESTION:
if the speed of a car doubles, why does the force of impact quadruple

ANSWER:
Here is that question that keeps coming up again and again: when one thing going some speed hits some other thing at rest, how much force is exerted? THERE IS NO WAY TO KNOW FROM THAT INFORMATION. See my FAQ page, first two questions. If two cars crash into an immovable wall, each loses all its kinetic energy. Kinetic energy is
½mv2, so the energy converted into heat and sound is 4 times bigger if the speed is 2 times bigger. If each stops in the same distance, the average forces will differ by a factor of 4, but you would expect the faster car to "crunch up" farther, not the same, so the average force would probably be less than 4 times bigger.


QUESTION:
I am planning on using a strain gauge to measure force (N) exerted over time for a shark captured on a long line. Is there any way I can convert these values to joules of energy used given that it is a static pull i.e. no distance travelled?

ANSWER:
If I understand, you are not hauling this shark in, he is just pulling trying to get away but failing. He is consuming lots of energy, but your line could just be tied to the pier and therefore does no work. Therefore, the force will not give you any information about energy because, as you note there is no distance. It takes no energy to hold the shark. The shark herself is doing work against the water and consuming energy and you cannot measure that by measuring how hard she is pulling.


QUESTION:
if a plane is flying 240 m/s in the air will the pilot hear the doppler effect?

ANSWER:
Yes, but it is not the usual. Usually we think of the Doppler effect as being due to either the source or the observer moving. In this case, the source (engine) and observer (pilot) are at rest relative to each other. However, the medium through which the sound moves is moving at a rate of 240 m/s, not so far from the speed of sound itself in still air.


QUESTION:
When an atomic explosion takes place, matter converted turned into energy. Is the reduction in mass accounted for by a quantified reduction in the total number of subatomic particles?

ANSWER:
No, the change in mass can occur even if the total number of particles remains constant.

FOLLOWUP QUESTION:
This sounds very strange and interesting! If what you say is true, then I am wondering where is the mass coming from? Do the particles themselves reduce mass?

ANSWER:
This is just E=mc2. Any nucleus weighs less than the sum of its parts. Here is how you understand that. Does it take energy to pull one proton or neutron out of a nucleus? Of course it does because it is bound in there and you must do work to get it out, thereby increasing the energy of the system. So it takes a whole bunch of work to disassemble a nucleus into its constituent protons and neutrons. Where does that energy go? Into mass, the mass of all the constituents is greater by an amount W/c2 where W is the work you did. So, if a heavy nucleus splits (fission, like in a conventional bomb) or if very light nuclei fuse (fusion, like in a star or in an H bomb), the mass after the fission or fusion is less than before since energy is released. It is subtle to understand why, but I have explained it in an earlier answer.


QUESTION:
This question came up in class and we went over it, but I still could not understand it. Assume the earth had sufficient nutrients and adequate water, no toxins present, uninterrupted growth, division every 20 minutes, and had e coli bacterium. How long would it take for this mass of bacterium to equal the weight of the earth? Is it possible?

ANSWER:
I think your teacher was trying to illustrate exponential growth. If you have not had any calculus, you will not understand my answer. Anything whose population changes in a way that is proportional to the population itself gives rise to exponential growth or decay. So, dN/dt=cN where N is the number at some time t, dN/dt is the rate at which N is changing, and c is a constant of proportionality. If c>0, N is increasing (growth); if c<0, N is decreasing (decay). This should make sense to you—if you double the number, you double the rate at which the number is growing. The solution to this equation is N=N0ect where N0 is the number at t=0. In our case we will choose N0=1 (start with a single bacterium). What we know is that if t=20 min, N=2, so 2=e20c. Solving this, c=20/ln2 where ln is the natural logarithm. Now, denoting the mass of one bacterium as m, we know that the mass M at any time must be M=Nm=me20ct and if we let M=Mearth, and solve for t we find t=20 ln(Mearth/m)/ln2. I will let you do the math and look up the constants. I get a little less than 2 days.


QUESTION:
I am attempting to build a 20 foot tall pole for use in a play. It will have minimal weight at the top of it, maybe 1 lb (a small flag). I am planning on using 2 10 foot long sections of Schedule 40 PVC pipe with a coupling to make the pole. I'm curious how large in diameter and how heavy my base should be. I was considering making a form 4 feet in diameter out of plywood and 2x4s and filling with about 200 lb of concrete. I want to be sure it is safe so if there are any drafts at the top or if someone happens to bump into it at the base that it won't fall over and hurt anyone. I believe the 4 inch Schedule 40 PVC pipes are approximately 10 lbs each, so the whole pole should only be about 20 lbs. Any recommendations on the base design or where I could find more information would be appreciated. I would like to minimize the base as much as possible without sacrificing the safety of anyone on stage.

ANSWER:
With a 4' diameter base, if somebody pushed on the pole at a distance of 4' above the ground, I figure that to push it just to the point of starting to tip it over would require a force of F=W/2 where W is the weight of pole plus base. The way I got this was to sum torques about the bottom of the pole, Fh-Ns=0, where h is the distance above the ground where F is applied, N is the normal force up from the floor (equal to the weight W), and s is the distance from the center to the edge where N is assumed to act if it is about to tip. So, for your design, a 110 lb force would be required and would have to be exerted until it had reached about 110 of tilt before it actually fell over. It seems very unlikely that somebody bumping into it would tip it. To be safe though, couldn't you find a way to attach the base to the floor?


QUESTION:
This question has been bothering me lately and hopefully you can answer. If you're in a car and the car is moving forward and if you would throw an item in the air why doesn't it go flying backwards considering it is not attached to the car or the people in the car?

ANSWER:
Actually, it does not keep perfect pace with the car
—it gets pushed back by the wind which you can feel if you stick your hand out the window. The effect depends on what you drop. If you drop a small, heavy marble, a passenger in the car would see it drop straight down, approximately. If you drop a cotton ball, I think you would agree that it would not appear to drop straight down.


QUESTION:N:
I was wondering what happens to massless particles - are they subject to gravity/able to to be acted upon by forces? If so, do they travel at a diffent velocity to particles with mass?

ANSWER:
The only known massless particles are photons and they, like any other massless particles that might exist, travel with the speed of light (they are light). We used to think that neutrinos were massless, but they are now known to have a very tiny mass, much smaller than other elementary particles. Regarding the effect of gravity on photons, yes, photons are bent by gravity even though they have no mass. See my FAQ page.


QUESTION:
What would happen if one split a radium molecule? This question is from my ten year old daughter's who has been reading about/admiring Marie Curie for the past two years.

ANSWER:
I think you must mean split a radium nucleus. Indeed, if you were to induce a fission of radium, energy would be released just like it is from any other fission of a heavy nucleus (like 235U or 239Pu), but you could not make a reactor with radium as a fuel because it is not fissile which means that it can be induced to fission by causing it to absorb slow neutrons and to therefore sustain a chain reaction. The energetics are such that if it were split into two fragments, energy would certainly be released. You can tell your daughter that I had the pleasure of working with Marie Curie's grandaughter Mme. Hélène Langevin-Joliot, also a nuclear physicist, at Saclay National Laboratory outside Paris in the 1990s.


QUESTION:
Why does torque depend on lever-arm distance? I don't really understand the logic behind it. Please don't use ideas of work,energy,moment of inertia etc to explain this. I just want a logical explanation for this using ideas from linear dynamics.

ANSWER:
In translational physics, force is a measure of how effective a push or pull is at getting an object moving (accelerating it). In rotational physics, torque is a measure at how effective a push or pull is at getting an object rotating (angular acceleration). So, think of opening a door. Suppose that you push the door out where the door knob is; your lever arm is large and this push is very effective in opening the door. Now, suppose you push with the identical force one centimeter from the hinges; the same force with a small lever arm is much less effective in opening the door.


QUESTION:
What should happen in the following experiment ... If you keep the height and length of a ramp the same and only vary the mass of the car by adding washers to add mass, should the car travel a lesser or greater difference? Some sources I've read say that the acceleration should decrease with an increase in mass, which means the distance traveled should also decrease. However, other sources say that the momentum will increase with the added mass causing it to travel further.

ANSWER:
Simple physics says there should be no difference. However, usually the heavier of otherwise identical cars wins. I have discussed this extensively in earlier answers which are linked to in my FAQ page.


QUESTION:
Does a theory eventually turn into a law or are they entirely separate? I have talked to many people about this subject and they have given me different answers.

ANSWER:
Usage by physicists of the words law and theory are ambiguous and vague at best. Newton's second law in its usual form, F=Ma, for example, is untrue for speeds not small compared to the speed of light; the theory of special relativity has never failed to predict the behavior of nature. See an earlier answer.


QUESTION:
My 13 year old daughter is studying physics and has asked a question I have no idea how to answer so I am hoping you can help. She has been told at school the energy of a wave is proportional to the square of it's amplitude and therefore independent of frequency. However she told me she has also read that energy of a wave is given by E=hf (E is energy f is frequency h is Plank's constant) and so is proportional to frequency! She is utterly confused by this and asks which one is correct?

ANSWER:
Your daughter has encountered one of the most important innovations of 20th century science
—quantum mechanics. Light is both a wave and a particle (called a photon). If you devise an experiment to prove that light is a wave (e.g. Young's double slit experiment), you will succeed. If you devise an experiment to prove that light is a particle (e.g. the photoelectric effect), you will succeed! Both energy statements in your question are correct. But they are not really contradictory. The intensity of light is the energy per unit time through an area, Joules/second/square meter=Watts/meter2. If you look at this as waves, this is proportional to the amplitude squared. If you look at it as particles, this is proportional to the number of photons passing through the area. Therefore, red light of some intensity has more photons than blue light of the same intensity because each "red photon" has less energy than each "blue photon".


QUESTION:
What is the wavelength range of the radiation from the sun that actually hits the ground?

ANSWER:


QUESTION:
if 2 lines start on the same point. And they are separated by a 1 degree difference in their heading. How far apart will they be extended out 365 inches. What I'm trying to illustrate is 2 people walking together in marriage - barely different - how far will they be apart in 1 year (365 days) because of 1 degree of separation.

ANSWER:
To most easily do this, convert 10 to radian measure, 10=
π/180=0.0175 radians. Since the angle is 0.0175=s/r (see figure at the right), s=365x0.0175=6.37 inches.


QUESTION:
I'm attempting to expand my understanding of some fundamental principles of physics, such as the laws of motion and energy. I've run into a bit of a puzzle dealing with the conversion and conservation of energy in reference to moving objects. I understand the conversion of energy in, say, a moving car: chemical (potential) energy of the gas is converted to kinetic energy, and this in turn is converted to heat energy when the breaks are applied. Where I get hung up is when energy is used to slow a moving object. Using the car as an example, rather than breaking one might downshift, thus using the engine to slow the car. This is where my understanding breaks down. It seems *more* chemical energy is being converted and put into the car to slow it down. What is the sum of this kinetic energy being converted to?

ANSWER:
Lots of the energy your engine generates is lost to friction. If you drive down a straight, level road, you are constantly consuming the chemical energy in the gasoline just to keep going. This energy all ends up as heat or sound in compliance with energy conservation. If you take your foot off the gas, you are not providing enough energy to keep going and you therefore must slow down, again because of energy conservation. If you downshift and take your foot off the gas, your engine goes faster but the gas you are providing is the same so friction in the engine is increased while gas consumption is not, so you slow down faster still.

QUESTION:
A better example may be a rocket in space. Let's assume it's not acted upon by gravity. The rocket engines fire, converting chemical energy into kinetic energy. Now, to come to a stop again, the rocket must turn around and again fire its rockets - and this is where I get confused. Though it "naturally" makes sense that this would "decelerate" the rocket, I'm confused about the principle of conservation of energy. It seems that in both instances of firing the rockets, kinetic energy has been put *into* the rocket. What is this kinetic energy being converted to when it is decelerated?

ANSWER:
This is not really the same question at all, so I will answer it separately. It is conceptually easier because there is no energy being lost to friction. The way a rocket works is that chemical energy is used to propel mass out the back (or front) and the rest of the rocket recoils; it is like the recoil of a rifle. Here is my simple model of a rocket: it is a 2 kg rocket with 1 kg of fuel. Now, the fuel is ejected so that the rocket recoils with a speed of 1 m/s; what is the speed of the ejected fuel? This is determined by momentum conservation. Linear momentum is the mass times the velocity, so before the rocket "burns", the momentum is zero and afterward it is (2 kg)x(1 m/s)-(1 kg)V where V is the speed of the ejected "fuel"; the minus sign is because they move in opposite directions. Conserving momentum (equating old with new), we see that V=2 m/s. Now, what is the chemical energy consumed? It is the change in kinetic energy
½mv2 before and after; originally, kinetic energy was zero and afterwards it was ½x2x12+½x1x22=3 J which is positive, energy was consumed. Now, consider braking. Suppose the 3 kg rocket+fuel has a speed of 1 m/s; so, its momentum is 3 kg m/s and its kinetic energy is 1.5 J. Now, the fuel is burned so that the 2 kg rocket comes to rest and the ejected fuel has a speed V. Again, momentum conservation gives us V; in this case V=3 m/s. So now the energy of the system is ½x1x32=4.5 J. So the change in energy of the system is again positive and equal to 3 J. The consumed chemical energy all went into kinetic energy of the fuel whereas in the acceleration situation, part went into the rocket and part to the fuel.


QUESTION:
What kind of force would it take to get a 1 or 5 or 10 gram object out of earth's orbit? What amount of force would be necessary to guarantee that the object leave the solar system? Or the Galaxy?

ANSWER:
To lift an object, you need to exert a force just a tiny amount bigger than the object's own weight. So, if you had a 10 lb object and wanted to move it away from the surface of the earth, first exert a force just greater than 10 lb and it will start moving upward. Once you have it moving, you can reduce the force to 10 lb and it will continue moving up with a speed of whatever you had given it. But, as you get farther and farther away from the earth, its weight gets smaller (see the answer right after this one), so it will start speeding up. So, now keep adjusting the force to keep it going with constant speed. So you see that the force required to move it far away is very small.


QUESTION:
Does every particle inside an object have a gravitational pull form all the objects in the universe. So each individual particle would have a pull of 9.81 ms-1 or sorry acceleration caused by some sort of mystical pull (any help appreciated).

ANSWER:
Two objects of mass m1 and m2 separated by a distance r exert a force on each other of F=6.67x10-11xm1m2/r2. For an object of mass M at the surface of the earth this just happens to be F=9.81M. For that same object at an altitude equal to the radius of the earth above the surface, that force would be four times smaller. In principle, all objects in the universe exert forces on each other, but gravity is only significant for very large masses and/or relatively small distances.


QUESTION::
I have read somewhere that hot water freezes quicker than cold water. if it is so, then please explain me the reason behind this.

ANSWER:
See an earlier answer.


QUESTION:
I know that freshwater can not be heated above 212 at sea level. What about freezing? Can freshwater attain a temperature below 32 degrees? If so, is there a theoretical minimum?

ANSWER:
At very high pressures, liquid water can exist at about -25 0C. See the water P-T diagram on Wikepedia. This happens at a pressure of about 3000 atmospheres.


QUESTION:
I just learned about the conservation of momentum in collisions in school. I wanted to check this in real life so I took some high speed footage of a friend hitting a tennis ball and after checking the frames, I found out that the momentum of the system before the racket hit the ball and after is not the same. It differs by about 50% and it occurs in all of the trials I had (3 in total). Is momentum not really conserved in real life?

ANSWER:
First of all, you need to understand under what conditions linear momentum is conserved. Total momentum of an isolated system is always conserved*.
 Now, if you just look at the tennis ball, the racquet exerts an average force on the ball over some time t so that the momentum should change by an amount Ft. So, if you measure the momentum of the ball before and after it is hit, it changes by an amount equal to Ft. But, do not forget that momentum is a vector and the ball changes direction, so if it has mass m and comes in with speed v1 and goes out with speed v2, the change in momentum is mv1+mv2. So, do not expect the momentum of the ball to be conserved. Suppose you look at the ball plus the racquet as the system; here momentum is not conserved either because your friend's arm is pushing on the racquet and therefore delivering an impulse during the collision. To see momentum conservation, imagine the racquet in the middle of space moving to the right with some speed and the ball moving to the left with some speed; after they collide the linear momentum will be the same as before because this is an isolated system. Regarding your final question, real life is seldom as simple as introductory physics would have it, but real life can often be understood to a good approximation.

*It can actually be a little broader than that since, to change the momentum of a system, an external force must deliver an impulse which is, essentially, some external force times the time over which it acts [I do not know your level, but impulse is F(t)dt].


QUESTION:
If a clock traveling at high speed runs slower than one fixed on earth does the same apply for computers processing information. Say two computers at the same rate are writing a copy of war and peace, one at a high velocity one on earth, would the one on earth be farther ahead. And if so could I, being on earth then tell the person at high velocity what will be printing out soon via radio signals ahead of the computer on their aircraft.

ANSWER:
Once again, I make reference to the diagram I use to explain the twin paradox. For a full explanation of this diagram, see earlier answers (1 & 2). In a nutshell, this shows the stay-at-home twin (computer in your question) moving up the time axis and sending out a light pulse once a year and the traveling computer traveling out to a distant star and back, also sending out a light pulse once a year. Think of each light pulse as being sent after turning 20 pages of War and Peace (pretty slow computers). Then, your idea is seen to fail because it takes time for the information to travel between the two computers. For example, the second pulse (40 pages read) sent out from the earth-based computer does not reach the traveling computer until it is turning its 120th page. In your question, the traveling computer never comes back. But, if it does come back, the traveling computer will begin getting advance information after he has received 8 signals (at which time he has also sent out 8). When the round-trip journey has finished, the earth-bound computer has read 400 pages while the traveling computer has read only 240.


QUESTION:
If I creat a huge (I mean massive) float system with a pully generator that will create energy from the rise and fall of the daily tide, will I eventually disrupt the tidal pattern? I've had the concept for years but am afraid of the outcome. Then I weigh in the damage we do daily other ways, and it would be as almost abundant as the sun. I am only worried about the tide. I realize I'd disrupt the waves.

ANSWER:
Tidal power is already used. You can get a pretty good overview from Wikepedia. Nothing you build will seriously disrupt the tides but it can have local environmental effects like riverbank erosion, disruption of ecosystems, etc.


QUESTION:
My son has already totaled one car in a curve, but he wasn't hurt. He thinks rigid suspension will solve problem, but he had rigid suspension in the car he totaled. I heard somewhere that when you reach a certain speed car can't handle curve. I guess the bank of the curve is involved. (I watch car races) I have limited math skills. But he would understand the math, and I would use it to convince him to slow down.

ANSWER:
When you round a curve a force is required because a car is accelerating when it turns even if its speed is constant. That is because the direction of your velocity is changing, just as truly an acceleration as when the magnitude (speed) of your velocity is changing. The force required to keep you moving with a speed v in a circular path of radius R is mv2/R where m is your mass. On a level road the only possible source of such a force is the frictional force between your wheels and the road. You can see that this is the case because on an icy road, which can provide almost no friction, it is almost impossible to turn at any speed. So, you can see that the faster you go the more frictional force you need from the tires to be able to negotiate the curve. In fact, since it depends on the square of the speed, doubling the speed quadruples the needed force. Now, the nature of static (not sliding) friction is that there is a maximum amount you can get which depends on the nature of the surfaces in contact (rubber and asphalt for this case) and how hard the surfaces are pressed together. In the case of a car with mass m, the weight is mg (where g is about 9.8 m/s2=32 ft/s2) and that is how hard the two are pressed together. The maximum force you can get is then Fmax=
μmg=mvmax2/R where μ is a constant determined by the nature of the surfaces (e.g. rubber on ice has a much smaller μ than rubber on asphalt). So, the fastest speed you can negotiate a curve on a flat road is vmax=√(μgR). For example, rubber on dry asphalt has μ≈0.6 so for a curve of radius 40 m, vmax≈√(0.6x9.8x40)=15.3 m/s=33.6 mph. If the road were wet, you would skid out at an even lower speed. If the road is banked the math is a bit more complicated and, you would find, you can go faster. Nevertheless, the suspension has nothing to do with the speed where you will lose traction with the ground. See also an earlier answer.


QUESTION:
Has the quantum concept "entanglement" been experimentally proven? Where might I find a layman's discussion of it?

ANSWER:
There have been numerous experimental verifications, the easiest to understand probably being entangled photons and their respective polarization states. There is a very nonmathematical discussion in Roger Penrose's book The Emperor's New Mind. While not mathematical, it is still pretty tricky to understand.


QUESTION:
Suppose we have tu push a large block on a very rough surface. It is assumed that the surface is so rough that we are unable to push it. We will say that the friction is acting in the opposite direction to our force and our force is not strong enough to overcome the force of friction. But when we are not pushing the block, then also the block remains stationary. There is no net force acting on the block. How can we say that the friction is acting only as long as we are pushing the block and ceases to act as soon as we withdraw our own force.

ANSWER:
If an object is at rest, the sum of all forces on it must add to zero. If static forces is one force then it will adjust to whatever the other forces are to maintain that zero sum. If your block is sitting on an incline, it does not go down to zero when you stop pushing on it; rather, it adjusts to equal the component of the objects weight which is trying to push it down the incline.


QUESTION:
If you are in a car and inside the car is a floating Helium balloon, if the car makes a hard turn, what direction does the balloon go?

ANSWER:
The opposite direction the car is turning. However, it might not be too pronounced because the air drag as it starts to move would restrain it. Of course, what it is really trying to do is keep moving in the direction the car was originally moving as the car turns, it just seems to you that it is being pushed out. A similar question may be seen in a previous answer.


QUESTION:
So if me and a friend are picking up a weight (30lbs) from either side lets say its flat like a board, we're each doing work to pick up the weight yes related to the angle and other factors but is it possible for each of us to be holding up 30lbs? Or is the weight always divided between the two of us, equaling less than 30lbs for each?

ANSWER:
If the board has its weight uniformly distributed and each of you is holding up one end, you each are exerting an upward force of 15 lb. If one of you is holding on closer to the middle than the other, he holds more than the guy at the end. But the sum of your two forces will always be 30 lb.


QUESTION:
While discussing atoms with some students we "briefly" discussed atoms and atomic structure being mostly empty space. The idea of solid matter being made of atoms is difficult to grasp for middle school. More so the idea that atoms do not touch. So of atoms never touch, why do I make sounds when I clap? Or create heat and friction when rubbing my hands? And how does fire affect items if atoms never touch? Even I couldn't answer these students questions!

ANSWER:
The key, I think, is that objects do not have to touch to exert forces on each other. Think of dropping a stone. Althought it is not touching anything it speeds up as it approaches the ground. Something is causing the stone to gain energy as it falls. Atoms (small size, about 10-10 m) are positively charged nuclei (much smaller size, about 10-15 m) with electrons on the outer parts of the atom. The electrons have a mass about 2000 times smaller than the nucleus, so you can see why atoms are often referred to as being mostly empty space. But, when two atoms come close to each other the negatively charged electrons in on atom see the electrons of the other atom and are repelled, that is they feel a force. This leads to friction. Fire is just rapidly moving atoms in a gas which then exert forces on atoms they are close to causing chemical reactions to occur (i.e. burning). Finally, I guess I do not like the statement that atoms do not "touch"; the electrons in one atom often overlap those of neighboring atoms.


QUESTION:
Hello, today i watched a programme about the universe. It stated that if a person in a space craft were to travel at incredibly fast speeds, by the time he got back to earth, being in space for a week, a huge ammount of time would have passed. I thought this was a facinating deduction, so i thought about it for a second and thought of this question: if the person in space could if possible, communicate to people on earth like on a webcam, or phone, what would happen? Would the people on earth speed up? Or what would happen?

ANSWER:
See an earlier answer.


QUESTION:
I have only ever seen depictions of the solar system with the same design: a series of planets in a line. But I've been wondering, all of these graphs basically depict the planets as orbiting more or less in the same plane (presumably with some wobble). So, is this true? Or is that for easy of orbit-distance-comparison? And if they are all in the same plane... why?

ANSWER:
All the planets orbit very nearly in the same plane. The reason is that the early solar system condensed from a big cloud of gas and dust which was spinning. As it got smaller it spun faster (conservation of angular momentum) and flattened out like a pancake. The planets formed by clumping of the matter in this pancake.


QUESTION:
Since the acceleration due to gravity of a planet like Earth can be given by the equation g=(GM)/r^2, where G is the universal gravitation constant, M is the mass of the body, and r is the distance from its center, what would happen to g very close to its center? The limit of that equation (assuming M is the mass of Earth for example) as r approaches 0 is infinity. How can you have an infinte amount of acceleration? What exactly would happen?

ANSWER:
The equation you quote is true only for point masses or outside of spherically symmetric mass distributions. It may also be applied inside of a spherically symmetric mass distribution with one important revision
M means only the mass inside of r. Therefore, the force at the center of the earth is zero. You might be interested in one of my FAQs.


QUESTION:
Assuming Einstien's theory, that matter is comprised of energy, that in fact, energy and matter are just two forms of each other, is correct; where does the energy in our bodies go, as we move throughout the day? how can we lose energy, without losing mass?

ANSWER:
If energy is used, mass is lost. But, think about it. The amount of energy in a mass is huge, so the amount of energy you use (which comes from chemical reactions) is so small that you could never hope to measure it. For example, the energy you would burn to climb a 9 m rope would be about 9000 J. This would correspond to a change of mass of about m=9x103/(3x108)2=10-13 kg.


QUESTION:
Is the net observation of either of two parties traveling toward one another at half of the speed of light equivalent to either one witnessing the other as traveling at the full speed of light, in that either one can be justified as "at rest" relative to the other one?

ANSWER:
You would think that, wouldn't you? Well, for very high speeds the intuitive ideas of velocity addition don't work. You should read an earlier answer for the details, but in the case you are imagining, each party will see the other approaching with a speed of 80% the speed of light.


QUESTION:
To get something off the ground you need to push the object with enough force against the ground (and atmosphere) to get it to overcome gravity and 'fly', but what would a theoretical space craft have to push against to get it to move in the vaccuum of space? I've always seen shows and movies pushing space crafts around with puffs of air and can't understand what the air is pushing against to make it move.

QUESTION:
How does a spaceship travel in space if the engines have nothing to push against?

ANSWER:
Both these questions illustrate a common misconception that the way a rocket works is by the exhaust pushing on something. The way a rocket works is that the rocket recoils when it "throws out" the exhaust. Imagnine being in empty space with a rifle; when you fire the rifle you and the rifle recoil. That is how a rocket works.


QUESTION:
What formula do you use to prove that a car can not jump abridge without a launch pad there to lift it off the ground? I know it can't happen but I don't know how to prove it!

ANSWER:
You don't need no stinkin' formula. You just need a simple qualitative argument. When an object is in free fall (meaning it is not in contact with anything, so after it has left the ramp), it can only do one thing
—fall. (I am assuming no effect from the air; an airplane obviously does not fall.) What "fall" means here depends on how it is moving:

  • If it is initially moving upwards (like off a launch pad), it will slow down.

  • If it is initially moving downwards, it will speed up.

  • If it is initially moving horizontally, it will speed up, downwards.

Therefore, if it is not originally moving upwards, it can never reach a point either level with or higher than where it was launched from.


QUESTION:
NASCAR "drafting" energy: If "drafting" behind another vehicle is lucrative enough in energy savings to merit the inherent danger it entails -- then recouping some of that energy by the vehicle generating it should be lucrative as well. In general, how much energy is available from a vehicle's "wake pressure gradient"? (my term) Could some of that pressure gradient potential be "tapped" with say a literal pipe-tap from the vehicle into the low pressure area behind it's own bumper? Could the resulting pressure differential be used mechanically in some way to result in a net energy savings for the vehicle itself? Or is the "drafting energy" only available to a disconnected/uncoupled vehicle?

ANSWER:
Aerodynamics is a very complex topic, usually best done with powerful computers and extensive wind tunnel measurments. Drafting is very interesting. It turns out that it is advantageous to both cars, but the trailing car gets the lion's share of the advantage. The reason has to do with the change in geometry of the two cars compared to the individual cars. So, it is not so simple as there being a source of energy in the wake of the leading car which the second car taps. You can be sure that the first car may not somehow extract energy from the turbulent wake behind it; there is no free lunch in the world of physics.


QUESTION:
i'm just wondering when we see the sky diving shows, why do the sky divers tilt downwards a lot?Does this have a effect on the forces acting on him? how?

ANSWER:
The air drag force is proportional to the cross sectional area presented to the onrushing air. If the sky diver orients himself in the direction she is moving, she goes faster; if she spreads out perpendicular to the direction she is moving, she slows down.


QUESTION:
a body is moving with a velocity 'v' with respect to (w.r.t.) a frame of reference s1.It bumps into a very heavy object and comes to rest instantaneously,its kinetic energy(1/2*m*v^2) as seen from the frame s1 is completely converted to thermal energy.Now a man moving with a uniform velocity 'V' (in the direction of the body) w.r.t. s1 observes the body , he notes that its initial kinetic energy of the body is 1/2*m*(v+V)^2 and that after it rams into the heavy body as 1/2*m*V^2 and concludes that the thermal energy produced is m*v*V+1/2*m*V^2.Which of the two answers is correct?

ANSWER:
So, to answer your question we just need to find out how much the kinetic energy changed as seen by both observers, right? Where you have gone wrong is that you have not taken the energy of the wall into account and you assume the wall has infinite mass. Of course, no wall has infinite mass but if yours did, it would have, for the moving observer, infinite kinetic energy both before and after the collision but those two infinities would not be the same! What you need to do is assume the wall has a mass M and the body has a mass m. So, your first observer sees a speed after the collision (conserving momentum) of u=mv/(M+m) and a change in kinetic energy of
ΔE=½(M+m)u2mv2mMv2/(M+m) which is the energy converted into thermal energy. In your second scenario, u=V+mv/(M+m) and ΔE=½(M+m)u2-(½mv2MV2)=½mMv2/(M+m), exactly the same. You can also note that as M-∞, ΔE-½mv2.


QUESTION:
Hey I've been reading some General Theory of Relativity and i understand that gravity warps space and time yet on earth what does gravity do does it warp anything.

ANSWER:
The warping of space-time is why you fall out of a tree. The warping of space-time is why relativistic corrections must be made to clocks in satellites so that GPS will work. It is not what "gravity does" it is what the mass of the earth does and it is what gravity is.


QUESTION:
Nuclear fusion reactions are kind of straightforward with the elecrotmagnetic force accelerating the two daugheter nuclei doing the large amount of work. Is the high speed of the neutron formed at nuclear fusion reactions due to an acceleration by the strong force, or is that reasoning too simplistic?

ANSWER:
You have it wrong, the electromagnetic force is not the source of fusion energy. If you imagine using an accelerator to bring two light nuclei together, that is just the "match" to ignite the fusion. The source of the energy is the fact that the mass of the fused nuclei is less than the mass of the unfused nuclei; mass is converted into energy. In fission, the source of the energy of that speedy neutron you are looking at is, again, from mass being converted into energy. When heavy nuclei split apart, the mass of the products is less than the mass of the original nucleus and thus energy is released. E=mc2. You might like to get more detail from an earlier answer.


QUESTION:
why does not moon escape toward the sun due to more powerful gravity of sun compared with earth?

ANSWER:
Is the moon not orbiting the sun? Just because it is in orbit around the earth does not mean that it is not also in orbit around the sun. The sun's gravity does not cause it to escape, just to follow the earth around the sun.


QUESTION:
We talked about this in my physics class but I was still confused. What are all the forces acting upon an elevator as it moves from rest upward and stops at its floor? To me the passengers would have something to do with the force, or am I wrong?

ANSWER:
You are right, the passengers have something to do with the force, but if you say the weight of the passengers is a force on the elevator, you are dead wrong. The weight of the passengers is a force on the passengers. The passengers exert a force down on the elevator, the magnitude of which depends on what the acceleration of the elevator is.


QUESTION:
If atoms are mostly empty space and if gold leaf can be as thin as one or two atoms thick, why can't we see straight through it? I understand in thicker objects there are atoms behind the atoms so you can't see straight through except that isn't the case for transparent objects like glass and diamonds. How can we see through them?

ANSWER:
First, gold leaf is nowhere near 1-2 atoms thick. It can be made a little less than 10-7 m which corresponds to about 1000 atoms thick. And, in fact you can see through gold leaf and the light passing through is bluish green. Gold is a metal and metals have conduction electrons which are more or less free to move around; the entering light has electric fields to which the electrons respond and absorb their energy. Transparent things like glass are insulators and therefore do not absorb much energy from the light. The detailed description of light transmission is very complicated. You might be interested in this site for information about colors of gold.


QUESTION:
OK, I'm a screenwriter endeavoring to get things right in my work. So, here goes:

  1. I am confused by what seems to me to be the mixing of Newtonian descriptive language with Einstein's whenever I hear someone talking about gravity. For example, the use of terms like "attraction" of gravity is really a Newtonian throwback, is it not? And, actually, isn't it incorrect as it is the deformation of space (G. Relativity) that causes the effect we refer to as"gravity"? Or, conversely (and my question), is it correct to purge Newton from my dialogue about gravity? If so, and I'm left with space, doesn't that mean, strictly speaking...(question 2):

  2. Is space gravity?

ANSWER:
This is a little tricky, because it involves semantics as well as physics. Newton was the first to appreciate the nature of gravity, its "action at a distance" nature. He was able to determine, by showing how the gravitational force of the sun could explain the motion of the planets (which was known from astronomical observations made by Brahe and systematized by Kepler), how gravity depended on distance r between attracting bodies (1/r2), and their masses (M1M2), and how strong the force was. This was an amazing achievement, but, it must be noted, it was purely empirical. It never addressed why or how the masses caused the "attraction". There is that word (attraction) that you suggest might be archaic. But, I would argue that regardless of whether we understand the theory behind the attraction or not, it is still clear to anybody that two bodies with mass attract each other. Einstein's amazing achievement was the theory of general relativity where the attraction was shown to be due to the warping of space-time, the why/how holy grail for gravity. It is just another step in the progression of science: data are gathered, the data are interpreted, an empirical theory for the interpreted data is proposed, and finally the reasons behind the empirical theory are proposed. I see no reason to "purge" Newtonian gravity terminology from a discussion of gravity, it is true and useful in many ways. I am quite sure that NASA does not use general relativity and the warping of space-time to calculate trajectories of space probes on their way to outlying planets. Any discussion of gravity should, in my opinion, not "purge" Newton (nor Kepler nor Brahe nor Copernicus, for that matter). I might add that general relativity is not regarded as the last word on gravity; there is no theory of gravity for very small distances, a theory of quantum gravity is sought but elusive. Hence, gravity stands alone and physicists seek a way of uniting it with the rest of physics. Finally, although I am not so mainstream here, I believe that the recent observations which are attributed to dark matter and dark energy may simply be indications that general relativity is not as complete and correct as is generally believed. The answer to your second question is that gravity is the warping of space-time by mass. I would urge you to have a look at my FAQ page for some earlier answers about general relativity.


QUESTION:
I've started doing year 12 work for next year and one of the topics is circular motion. One of the examples of a centripetal force in the text book is when a car turns along a curved track, and it says the centripetal force is friction. But i dont understand this, as i thought that friction is the resistance to motion and the wheels arent facing the 'centre' of the curved track, and thus how is the centripetal force friction? This also lead me to wonder how a car moves in physics terms, coz i also have heard that a car moves due to friction with the ground, which i also do not understand when friction is the force opposing motion. So if you could, can you also explain why cars move once the motor has provided a torque on the wheels, because im battling with this concept of friction especially.

ANSWER:
It is a mistake to think that friction always takes energy away. For example, if there is a box sitting on an incline and not sliding, what is keeping it there? It is the static friction between the box and the incline and it is not changing the energy at all; if there were no freiction, the box would slide down. The car going around a curve stays on its circular pathe because of friction. If there were no friction, it would be impossible to negotiate a curve, you would slide off (like when the road is icy). The wheels are not slipping, so, again, it is static friction because at any instant the tires are at rest relative to the road surface. The only acceleration of a car going around a curve with constant speed is toward the center of the circle, so that has to be the way friction acts. For a car to start moving, there must be friction (again static friction); if there were no friction, the engine would cause the wheel to turn but it would slip and the car would go nowhere. The friction must exert a force backward on the wheel in order for it to not slip. Therefore, the road exerts an equal and opposite (forward) force on the wheel. The friction force on the car by the road is the force which causes the car to have a forward acceleration. Once you are moving, there are lots of other kinds of friction which come into play
—air drag, friction from the wheel bearings, frictional drag from the engine and transmission—which all cause the car to want to slow down, and the forward force of the road on the wheels balances these forces when keeping the car going at constant speed.


QUESTION:
If a black hole pulls material in doesn't it gain mass and therefore slow down?

ANSWER:
Yes, it gains mass. But, it gains the mass by interacting with something else. So, to find out what the black hole is doing after the collision, you have to conserve momentum of the event when it gained that mass. It is pretty much like the collsion of two putty balls
—you start with two and end with one. For example, suppose the black hole is at rest and the to-be-captured object is far away and has a velocity v directly toward the black hole and a mass m. So the momentum of the system is mv. After the collision, the mass must still be mv (that is what momentum conservation means). So, if the mass of the black hole were 999m, the speed after the collision would be V=mv/(1000m)=v/1000. You could go through the whole thing for other initial conditions, for example if the black hole were moving with a speed v and the object were at rest, the speed after the collision would be V=999mv/(1001m)=0.998v, slower, as you suggested, but not really for the reason you suggested. (I have done this nonrelativistically, but it conveys the idea.)


QUESTION:
How much energy does CERN use to smash 2 protons together to try & see the Higgs particle? ie in laymans terms, would it be equal to lifting a chevy off the ground...or a train engine?

ANSWER:
There is a link from CERN which fully discusses the energy content of the LHC beams. The number they give for the beam at full intensity is 362 MJ. They show that this is the energy of a 3200 kg Subaru going about 1000 mph. It is the energy which could lift that car about 10,000 m, or about 30,000 feet. But maybe you are asking about just the energy of the two protons? That is much smaller, because the beam contains about 3x1014 protons and so the energy of each proton would be about 10-6 J, not enough to notice macroscopically.


QUESTION:
How would a helium balloon behave in the zero-gravity environment of the International Space Station, or any other orbiting spacecraft?

ANSWER:
The reason the balloon goes up is that the buoyant force is greater than its weight. The buoyant force results from there being a greater pressure on the bottom of the balloon than the top. There is a greater pressure on the bottom than the top because the pressure is the result of all the weight of the air above some point, and there is more air above the bottom than the top. So, if you have the international space station in empty space the balloon would neither rise nor fall because it has no weight and there is no buoyant force. In orbit, it is not really a zero-gravity environment but rather a free-fall environment. Still, the pressure difference, if any, inside the orbiting spacecraft would be probalby be too small to cause any buoyancy.


QUESTION:
I am a 4th grade teacher and one of my students asked me a great question....Why don't clouds move when a jet flies through them? I could not answer him. Can you help me?

ANSWER:
Well, the cloud certainly does move. The plane pushes the cloud out of the way to make room for itself, it creates currents and eddies around its surfaces which cause those parts of the cloud to violently move around. Think of similar situations: Does the water move when you dive into it? Does a watermelon move when you plunge a knife into it?


QUESTION:
What prevents velocity in a known direction from being measured by firing a photon in that direction and timing precisely how long it takes to travel a known distance? If the speed of light is in fact constant regardless of frame of reference, it would be possible to find the velocity by (known speed of light) - (observed speed of photon fired)?

ANSWER:
First of all, the speed of the photon is always the same, you would have to measure its momentum or energy before and after. But, the whole point of the example of bouncing a photon from something is to demonstrate the futility of trying make a precise measurement of the position or momentum of some particle because the act of measurement affects that which you are trying to measure. The heart of the matter, though, is that it is a fact of nature that some things are simply unknowable. You cannot know both the position and speed of a particle to arbitrary precision.


QUESTION:
Consider a water tank sitting on its stand and a pipe with an inner diameter of 6 inches was used to deliver water from the tank to the ground below it. The vertical height from where the pipe is joined to the tank and the ground below is 20 feet. However, rather than coming straight down, the pipe comes down at an angle of 45 degrees and is then joined at the bottom so that there is a 3 feet length of pipe running horizontally (180 degrees) along the flat ground. Neglecting friction loss in the the pipe and in the bends of the pipe, what will be the flow rate of the water coming out at the end of the pipe? How much energy will the water have? Is there a formula that can be used to work out the final flow rate and energy? Sorry if the scenario sounds silly. I can provide a sketch of the scenario if it helps.

ANSWER:
I need two more pieces of information: how deep is the water in the tank and what is the cross sectional area of the tank (if it is much bigger than the cross sectional area of the pipe, it can be neglected with little error.) The shape of the down pipe makes little difference, just how far it opens from the top surface of the tank. You need Bernoulli's equation which says that
½ρv2+ρgy+P=constant where ρ is the density of water, P is the pressure, v is the speed of the fluid, g is the acceleration due to gravity, y is the height. In your case the pressure at both the top of the tank and at the outlet of the pipe is the same (atmospheric), v at the top surface of the tank is approximately zero, and we can choose y=0 at the ground, so v=√(2gy). For example, if the distance from the outlet of the pipe vertically up to the surface of the water were 30 ft, the speed of water coming out the pipe would be v=√(2x32x30)=44 ft/s. The cross sectional area is πx0.252=0.2 ft2 and so the flow rate R would be R=44x0.2=8.8 ft3/s=66 gal/s. A general expression can be written for your case as R=12√(20+h) gal/s where h is the height of the surface of the water in the tank above the top of the pipe. For this to work, the speed of the drop in the tank must be small. You cannot say the water has energy, you can say each gallon of water has an energy Emv2 where m is the mass of a gallon of water. When you figure that out, the energy per second is P=ER which is power; if you do that calculation in SI rather than English units, it will come out in watts. For my example above, R=66 gal/s=0.25 m3/s, v=44 ft/s=13.4 m/s, E=½x1000x13.42=89,800 J/m3 (1000 kg is the mass of 1 m3 of water), so P=22.4 kW. That's a little surprising to me, but a 6 inch pipe is pretty big. And, of course, you would not be able to extract all this to use.


QUESTION:
How small would the human body be if you took out all the space between the atoms???

ANSWER:
See an earlier answer.


QUESTION:
Ok, so I got into a big discussion today with my physics Honors teacher and he didn't even say who was right or wrong. The problem was that there is what you could consider a clothesline with a cable (for purpose of the problem the weight of the cable does not matter) and a weight is suspended exactly in the middle of the cable. The weight of the object was 25 N. The angle at which the cable meets the object/block is 30 degrees. What is the tension in the cable? My theory behind this was that when you find the force tension it is for only half of the whole cable so you need to double the force tension and my teacher argued that you do not. So with this problem he said that the answer was 25 N of force tension and I said that it is 50 N of force tension. Who is right??

ANSWER:
I am sure that you will not be too surprised to find that your teacher is right. Maybe this would be easiest for you if you do not think of the tension in the cable but rather the tensions in the left and right parts of the cable. If the angles of the two parts of the cable were different, the tensions would be different. So, right away, the notion of doubling the tension to account for the two halves is seen to be faulty reasoning. If you call the two tensions TL and TR, you can see that the horizontal components of the two tensions must be the same: TLcos300-TRcos300=0, so TL=TR=T. Similarly, the vertical components of the two tensions must hold up the weight: TLsin300+TRsin300-25=0=T/2+T/2-25=T-25, so T=25 N. This means that the tension in each side of the cable is 25 N and you could say, in this case, the tension in the cable is 25 N everywhere.


QUESTION:
So I was doing my physics homework the other day and we had to figure out what the the horizontal and vertical velocity of a long jumper and the answer for the vertical was less than the acceleration from gravity. Anyway this got me wandering if you tried to launch an object upward with a speed lower than the acceleration due to gravity would it actually go anywhere?

ANSWER:
When you say "speed lower than the acceleration due to gravity", it makes no sense because speed and acceleration are different things. It is sort of like saying "a temperature greater than the color of a carrot".


QUESTION:
Many people place sandbags in the rear of their vehicles to increase winter traction. Many say only in rear wheel drive vehicles, generally over the rear axle or behind it (for more lever arm). Thusly weighted pick up trucks are said to benefit from better weight distribution correcting "inherent oversteer." I could find only a few naysayers who say that there's little or no advantage gained or that the sandbag can become a dangerous flying object. It seems to me that if there's enough weight to affect the handling of the vehicle then if that weight shifted, as one could expect it might during reactive emergency driving, then the handling would be, at best, momentarily unpredictable. So, specifically regarding a new rear wheel drive small pickup (e.g. a 2011 2WD Ford Ranger) with the all season fairly aggressive tires it came with, what is your analysis of the physics of adding weight to a pickup for winter traction? Aside from being perhaps unsafe does it really help in starting, stopping, cornering and/or hills?

ANSWER:
The force which drives your car forward, which allows you to stop, which allows you to turn and maneuver is friction. The friction you are able to get without your wheels slipping is proportional to the weight of your vehicle. So, it makes sense that you should increase the weight of your vehicle, right? Well, the force which you need to accelerate, brake, turn or maneuver is also proportional to the weight of your vehicle. So, according to elementary physics, there is no advantage to additional weight. However, since a vehicle is not a point mass and has four wheels, changing the relative loads on the rear and front wheels may make a difference. It can be a complicated problem in the real world, but based on elementary physics, I would guess that any gain in carrying sandbags is minimal.


QUESTION:
Basically what im trying to determin is the amount of restraint required to hold an item on an incline. In my occupaton im required to restrain various items for predetermined forces. Ie 3Gs forward, 2Gs vertical. However in this case I am not given the amount. Normally I would take the weight of the object, multiply it by the force it might incur then divide it by the rated value of the restraint being used. What I want to know is, if an object is on a 10 degree ramp, how much force is required to hold it. The way I figure is if the ramp were at 90 degrees it would be 1G and if the ramp was at 0 degrees it would be zero. So if the ramp was at a 45 degree angle would it not be 0.5 Gs. So 10 degrees is .111 20,000 lbs vehicle multiplied by .111Gs is 2220 divided by 5000lbs (rated strap strength) = 0.444 thus requiring only 1 strap I know that there are other factors being ignored like the coefficient of friction between the ramp and the pneumatic tires of the vehicle. (in this case 0.030 ) Also the strength of the strap being affected by the angle that it is applied. (we will say it is getting 100 percent of its rated strength)

ANSWER:
You are on the right track, but the force is not linear from 00 to 900 as you are assuming. Ignoring frictional forces, the component of the weight parallel to the incline is the weight times the sine of the angle, so sin450=0.71 and sin100=0.17. So, your 20,000 lb vehicle would require a strap capable of holding 20,000x0.17=3400 lb on a 100 incline.


QUESTION:
As I uderstand it, Carbon 14 is formed by neutrons comming from cosmic rays witch interract with Nitrogen(7 protons + 7 neutrons). Nitrogen lose a proton and win a neutron wich become carbon 14 (6 protons + 8 neutrons), right ? Then, the part that I do not get is how Carbon 14 decay to Nitrogen. Wikipedia says that Carbon 14 loses eletrons and electrons antineutrinos to become Nitrogen again, but I don't understand because Nitrogen is 7 protons. Is Carbon gaining a proton ? Can you make this more clear for me ?

ANSWER:
The 14C decay is what is called beta decay. Inside the 14C nucleus one of the neutrons spontaneously turns into a proton, an electron, and an antineutrino. The electron and the antineutrino leave and the net effect has been to change 14C to 14N.


QUESTION:
Why is that if you push a book slowly off the edge of a table it will rotate, but it you give it a good, hard shove, it just flies forward and doesn't rotate much?

ANSWER:
As soon as the center of gravity of the book is beyond the edge of the table, there is a net torque which will cause an angular acceleration around the axis along the edge of the table. If you push the book slowly, there is a long time for the acceleration to act so it will acquire an appreciable angular velocity. The hard shove gives the torque very little time to act giving only a very small angular velocity. Once the book is in the air, there is no torque on it.


QUESTION:
how does lightning conductor works during a thunder storm ?

ANSWER:
The basic idea is to provide a conducting path to take the current from the lightning and send it to the ground. However, it also "attracts" the lightening so that it hits the rod rather than the house. When a charged cloud passes over, the objects under it become charged by induction. The rod has a sharp point which means when the rod becomes charged, the electric field near the tip becomes very strong and often strong enough to cause electrons to stream into the air; this is called corona discharge (or St. Elmo's fire if on the top of the mast of a ship). This charge density and strong electric field then has the effect of directing a strike at the rod.


QUESTION:
Do protons generate magnetic field around them ?

ANSWER:
Any electric charge which is moving creates a magnetic field. Since a proton has charage, it will generate a magnetic field when moving. However, a proton also has a magnetic moment, that is it looks like a tiny bar magnet. Therefore a proton has a magnetic field even if it is sitting still.


QUESTION:
Our state is considering the allowance for an increase in tractor-trailer weights from approximately 80 to 100 thousand pounds. The idea is that consumers will benefit because a given truck can haul more goods at a lower fuel cost. Law enforcement officials are against the measure because of the increased dangers due to increased stopping distances of the heavier vehicles. I'm curious about the relationship between increased momentum and stopping distances of these big rigs but don't have a good sense of how to get at it with a calculation. Would this comparison be better studied by actual measurements?

ANSWER:
According to the first approximation, which works very well, the weight of a vehicle has no effect on its stopping distance. See a recent answer. The reason is that the frictional force between the tires and the road is proportional to the weight of the vehicle and acceleration, which determines the stopping distance, is inversely proportional to the weight; the weight does not matter. However, the braking system can also be a factor because the brakes, ideally, just keep the wheels from skidding (what antilock brakes do automatically) and if they are overburdened (for example, overheat) they can fail. Yes, physics is an experimental science and doing actual measurements would be the best way to determine the answer.


QUESTION:
Is a golf ball going fastest after it leaves the club head striking it, or is there still some accelertion as it overcomes inertia?

ANSWER:
Nothing "overcomes inertia" on its own. Only a force causes an acceleration and acceleration is what is meant by overcoming inertia. The ball has its largest speed as it leaves the club because that force ends then. When the ball is in flight, only two forces act on it
—air drag which always slows it down and gravity which slows it down on the way up and speeds it up on the way down. But gravity can never speed the ball up more on the way down than it slows it down on the way up.

FOLLOWUP QUESTION:
Sorry I didn't do a better job asking the question. My family and I are not looking to be told when peak acceleration happens, we want to prove it mathematically. We have put together a derivation to show that at its peak velocity a golf ball is still in contact with the club head. Starting out with the length of the drive we use the formula for projectile motion to find the peak velocity of the ball. We work backward from there to derive all of our unknowns but we run into trouble when we use the formula for uniform motion with constant acceleration to calculate how far the ball and the club travel during the time to accelerate. Because the distance the club travels is expressed as S = u * t and the distance the ball travels is express as s = u * t + a * t^2 / 2, there is always a few millimeters distance between the ball and club when the ball reaches peak velocity.

ANSWER:
There is a difference between peak velocity and peak acceleration. For example, a pendulum has its maximum velocity at the bottom when the acceleration is zero. What you are doing is wrong because the acceleration is not uniform as you assume. Let us, in accordance with my original answer, focus only on the time during which the club is in contact with the ball; the time after it leaves the club will always have a smaller velocity than when it left the club. The graph to the right shows what the force which the club exerts on the ball is likely to look like. The club first touches the ball at t1 and the ball leaves the club at t2. The force over the time of acceleration is not constant because the ball behaves like a spring. When the force is biggest, the acceleration is biggest because of Newton's second law, F=ma. However, the ball is speeding up over the whole time interval and will therefore be going fastest at t2. The area under the force curve is called the impulse and is equal to the change in momentum of the ball which, because the ball starts at rest, is mv where m is the mass and v is the speed at t2 of the ball. If you call Favg the average force on the ball, then mv=Favg(t2-t1). Putting in some numbers I found on the web, m=45 g, Favg=3000 lb, v=250 ft/s, I estimated that the time of contact is less than half a millisecond.


QUESTION:
During a discussion about basic physics I was having with my daughter, she asked an interesting question. She saw a story about helicopters that mentioned the helicopter blade tips could exceed the speed of sound creating the characteristic “whop, whop” sound, or small sonic booms. She wondered if it would be possible for the blade tips to ever exceed the speed of light. I was intrigued by this so I worked for a few hours to come up with an equation to answer her question. I am not a math expert, but using simple algebra I came up with the formula: r=(60V) / (2πR) where r = radius of the blades in meters, V = velocity in meters per second, and R = blade RPM. From this I calculated that the blade tips would exceed the speed of light if the blades had a radius of 286.28 kilometers and were spun at 10,000 RPM. This of course assumes the materials used could withstand the forces involved. My question: is this formula accurate and given the radius and RPM mentioned, what is the formula to calculate the forces along the blades starting from the center moving to the tips (centrifugal, centripetal?)?

ANSWER:
Nothing can exceed the speed of light. The qualitative way to explain why for this case is that a mass increases as the speed increases. As you go farther out on the blade, the mass increases more and more until you are, at the crucial length, requiring an infinite force to keep the mass moving in a circle. The speed of something a distance r from the end would be v=r
ω=9.55Rr where R is rpm as per your notation and ω is the angular velocity in radians/s; it looks like your equation got 60/(2π) where you should have had (2π)/60. So, I get the length would be just 3.14 km if it could work. Suppose the blade had a linear mass density λ kg/m and a length L. So, the mass of the whole blade would be Lλ. If you calculate the force classically, ignoring relativity, you would get F(x)=(L-x)λxω2 where x is the distance out to the point where you are calculating the force necessary to keep the mass beyond from flying off. But, relativistically, λ depends on x, λ=λ0/√[1-(ω2x2/c2)] where λ0 is what the mass density is if the blade is at rest. So, you see, if ωx=v=c, the mass of the whole blade becomes infinite meaning the force needed anywhere is infinite.


QUESTION:
If like 3 water tanks are connected by pipes so water can flow to equilibrium, will the water levels still equalize even if the tanks are sitting on springs? To me it seems yes but mathematically proving is difficult.

ANSWER:
Yes. The reason is easier than you think. The springs do not exert forces on the water, they exert forces on tanks. Therefore, when considering the forces on the water the springs do not enter. The springs would compress or uncompress as the amount of water in each tank changes, but it would be no different from what would happen if you pushed the tanks up and down.


QUESTION:
if you had an object that was the size of a grain of sand but its "mass" (might not be the best term) was the same as the the suns, would it be possible for this grain of sand to have its own planets revolve around it?

ANSWER:
Yes. What matters is the force which the "sun" exerts on the objects orbiting it and that is determined by the mass of the "sun", the mass of the objects, and how far they are from the center of the "sun".


QUESTION:
My car travels at 55 mph. Are the outer portions of my tires moving at a faster speed than my car? Im thinking it does, but have no formula for it.

ANSWER:
The point of your tire which is touching the ground is at rest. The axel of that wheel is going forward with a speed of 55 mph. The top of the tire is going forward with a speed of 110 mph.


QUESTION:
A vertical narrow tube has a photon emitter at one end, so that photons travel through the tube and are emitted along the verical axis of the tube (all in a frame moving left to right near the speed of light). Do the photons emitted from the end of the tube leave at an angle to the vertical, as seen by a non relativistic frame? If so, do the photons hit or move toward the wall of the tube on their way inside the tube?

ANSWER:
First, view from the frame of the tube. The photons travel straight up the center of the tube and never hit it. Therefore, all observers have to agree that the photons do not hit the tube. Certainly the photons, as seen by the stationary observer, do not move vertically. Rather, they have a component of their velocity which is v along the direction of the motion of the tube but their total velocity is still c. The angle they make with the vertical is therefore
θ=sin-1(v/c). What you are describing is very similar to the light clock.

FOLLOWUP QUESTION:
you said that both the moving observer and the stationary observer must agree on whether the photon hits the wall of the tube or not. You said the moving observer see the photon traveling straight up the tube. but you also said the stationary observer sees the photon traveling at an angle to the tube wall. you didn't say whether that angled path resulted in the photon hitting the wall and how this discrepancy is resolved.

ANSWER:
There is no mystery here
—the tube has the same horizontal velocity as the photon does, v. So, every time t that the photon moves a distance vt, the tube also moves a distance vt and so the photon stays in the middle of the tube.


QUESTION:
If we are having two balloons suspended with thread. When we blow air between them why they attract each otherhence having like charges in both the balloon?

ANSWER:
Bernouli's equation says that in a fluid the pressure gets smaller as the velocity gets bigger. So, the pressure between the two balloons becomes smaller than atmospheric pressure (which is pressing on the outer part) and they go together. This is the same kind of force which helps airplanes fly and causes spinning balls to curve.


QUESTION:
if you fall 10 feet from a hayloft, what part of your body would hit the ground first?

ANSWER:
There is no way to answer this question. The orientation of the falling body at any point is determined by the initial conditions. For example, just jumping would result in feet first, diving head first would result in head hitting first. It is also possible to change the answer by what you do as you fall. For example, a diver can make his body spin faster by doubling into a ball and slow the spinning by straightening back up; and a cat can land on its feet.


QUESTION:
If my car weighs 5600 lbs and I'm traveling 35 mph. How long will it take to stop under normal road conditions?

ANSWER:
The weight of the car does not make any difference. The coefficient of friction between rubber and dry asphalt can be anywhere from 0.4 to 0.8, from which I calculate stopping times of 2-4 s and corresponding stopping distances of 50-100 ft.


QUESTION:
What is the diffrences between atomic bomb and nuclear bomb. Can both weapons be consider nuclear weapons

ANSWER:
Atomic bomb is an archaic name. It is what nuclear bombs were initially called the public was not so well informed about physics. Really, a conventional bomb should be called an atomic bomb since it gets its energy from chemistry which is energy from atomic forces. Nuclear bombs (fission and fusion are the two kinds) get their energy from the nucleus. But, atomic bomb is a name which survives and is just an old-fashioned name for nuclear weapons.


QUESTION:
Does the amount mass of an object really not effect the gain of speed withing a field of gravity? FE: DO a large meteor and a small metal BB fall at the same speed exact speed, If the factor of wind resistance and friction are taken out? And do you have a formula for this? please keep in mind that this is not for homework but to clarify my curiosity.

ANSWER:
Constant acceleration due to gravity is an idealization which assumes a uniform graviational field which means that the force is always straight down, the earth is perfectly flat, the force is independent of how far you are from the earth, the earth is not rotating, and that the earth has infinite mass. These are all literally false, but for most everyday problems they are an excellent approximation. If, as you stipulate, there is no air drag, then you would be hard pressed to measure any difference in the accelerations of a 10 ton meteor and a BB. If the altitude is very small compared to the radius of the earth and if the mass is very small compared to the mass of the earth, acceleration is constant.


QUESTION:
What would the transit time between Earth and Neptune be (pick your own planetary positions) assuming an acceleration/deceleration curve that maintained 1 g of force on the spacecraft for the entire journey? In other words if our hypothetical spaceship wanted to simulate gravity by means of constant acceleration as well as deceleration (for orbital capture) how long would it take to get to Neptune. I realize that one would have to take into account the trajectory of the spacecraft since getting from one planet to another is never a straight line, but any rough answer would be much appreciated.

ANSWER:
A constant acceleration of g
≈10 m/s2 to the halfway point will take the same time as the deceleration the rest of the way, so we can just calculate the time for the first half of the trip and double it. The distance from earth to Neptune is on the order of 4x1012 m. Starting from rest, the position is x=5t2=2x1012, so t=6.3x105 s. This calculation is correct only if the maximum speed is much smaller than the speed of light; the biggest speed is v=10t=6.3x106 m/s which is still about 50 times smaller than the speed of light, 3x108 m/s. So, the time would be about 1.3x106 s≈15 days. Pretty speedy!


QUESTION:
If you are holding a 3 pound, 2'x2' sign in a 5 mph wind how many pounds of resistance would you be holding?

ANSWER:
The weight makes no difference. The force on the sign can be roughly approximated as
FAv2 where A is the area and v is the wind velocity, but these must be expressed in SI units, A in m2 and v in m/s and then F is in Newtons. Doing the conversions for 4 ft2 and 5 mph and then converting Newtons to pounds, I find that the net force from the wind is about 0.1 pounds. 5 mph is a pretty gentle breeze.


QUESTION:
During beta decay, a neutron splits into a proton and an electron inside the nucleus. The electron is released from the atom which is the beta particle. I have studied that the atomic number of an atom increases by 1 after a beta particle is emitted. But the daughter atom has 1 proton in extra to that of the number of electron. So why does not the atom carries a positive charge of 1 unit?

ANSWER:
Immediately after the decay, the daughter atom is missing one electron. However, the electron which escaped ensured that charge was conserved. What will happen if there is a positive ion is that it will rather quickly find an electron to capture and become neutral, maybe even the beta particle that came out.


QUESTION:
Is quantum physics a theory or proven science? I have asserted that it is a theory, in debate with a friend, who demands that it is a complete proven science. With the exception of some experimentation, I can find no fqctual evidence with finished proof.

ANSWER:
It would be my opinion that there is no such thing as a "complete proven science" and no way you can provide a "finished proof" for any theory. It is really a question of semantics and I do not know how philosophers of science define theory, so I am not speaking authoritatively here. However, it seems arrogant to me to claim that any theory is incapable of being found incorrect in some limit, just as Newtonian physics, amazingly successful in its own realm for hundreds of years, was later to be found to be only an excellent approximation of the theory of special relativity at low speeds. To answer the question specifically about quantum mechanics, I heard a statement on a Nova show the other night that there has never been an experiment which was not described by quantum mechanics; that's about as close as it gets to "proven science". Who knows what future experiments might hold, though?


QUESTION:
I wonder, if we had a possibillity to accellerate a manned vehicle constantly up to, let's say, mach 5 or faster how slow/fast would the accelleration have to be to keep the human from dying? Would the pure speed itself kill the human? And would there be a theoretical way to construct a chamber that completely isolates it's contents from the force that sudden acceleration or something like dropping it would produce?

ANSWER:
Manned spacecraft have speeds far larger than mach 5. The maximum amount of acceleration which a person can endure depends on the duration, the person, training, etc. If you travel with constant velocity, you feel exactly the same as if you were at rest no matter how big the velocity is. Finally, there is no way to construct your hypothetical chamber.


QUESTION:
I have struggled to get the simple explanation (non-mathematical form) for my troubling question from books. Still no luck. Whts is force ? (as a concept) What is the relationship between energy and force? I am not a physicist so could you please enlighten me and explain the concept in layman's term for me? Or with some analogy?

ANSWER:
A force is a push or a pull. It is one of those things you need to have an intuitive feel for to begin doing physics. It is that which causes an object to accelerate. In physics accelerate can mean two different things, either change speed (speed up or slow down) or change direction. Energy is a more subtle concept. To understand what energy is you need to understand the concept of work. In physics, work is done by a force only if the force is exerted over some distance. For example, if you push with a force of 1000 pounds against an unmoveable wall, you do no work, but if you push a box across the floor with a force of 10 pounds, you do work. When you do work on an object you increase its energy. So, when you push the box across the floor, it acquires what we call kinetic energy, energy by virtue of motion, because it started at rest and ended up moving. If you use a force to lift an object so that it starts out and ends up at rest, you have still done work but now the energy you have given it is called potential energy, energy by virtue of position. Different kinds of energy can transform to others. For example, if you drop something its potential energy is converted into kinetic energy as it falls.


QUESTION:
My question is, the moment of inertia for a ring and a hollow cylinder about d same given axis is the same i.e. MR^2. its conceivable mentally, that being a rigid body, i only need to rotate one out of many rings that totally make up a cylinder, so or the stretch rule, as stated in wikipedia http://en.wikipedia.org/wiki/Stretch_rule, bt what is the mathematical proof for the rule???

ANSWER:
I never heard of the "stretch rule". Start with an object with infinitesmal thickness dz and moment of inertia dI1=
[∫∫r2ρ(r,φ,z)rdrdφ]dz=[∫∫r3ρ(r,φ,z)drdφ]dz about some axis. Now, add another "slice" with the same shape and the same r,φ dependence of dI2=[∫∫r3ρ(r,φ,z)drdφ]dz. Keep adding new layers to build up the whole object. For every slice, the integral over r,φ is just some function of z, f(z). So, finally, I=f(z)dz.


QUESTION:
enrico fermi once pointed out that a standerd lecture period [50 mint] is close to one micro centuery. how long is a micro centuery in minutes. and what is the percentage difference from fermi claim?

ANSWER:
Well, this isn't physics, but 1
μCentury=(100 yr)(365.25 days/yr)(24 hr/day)(60 min/hr)x10-66=52.6 min. Fermi was a real smart guy, particularly famous for "back of the envelope" estimates.


QUESTION:
I remember from an old schoolboy science book that if one were to remove all the 'space' from the atoms in a human being, the collective mass would sit on a pin-head. And, if it were done for the whole human race, we would be nothing more than a super dense sugar-cube. I also read that if this were done for the Earth, and all the 'space' was removed from it's atoms, it would be about the size of a rugby stadium (about 500m across). Would you concur with this? And contniuing this theme, what would be the size of the Sun?

ANSWER:
Basically, you are asking what is the mass density of the nucleus because the rest of an atom has negligible mass and the electrons can be taken as having no volume. This number is about 2x1017 kg/m3. So, take a 90 kg man with a density of about 1000 kg/m3 and a volume of 90/1000=9x10-2 m3. His new volume would be about V=90/2x1017=4.5x10-16 m3; so his size would be on the order of the cube root of this number, 8x10-6 m=8x10-3 mm, about 10 microns. I am not going to do any more, you do any others you are interested in.


QUESTION:
I know that a particle gains mass as it approaches the speed of light, but why? What actually happens to it? Can you explain in laymen’s terms how this is possible? I can imagine a large object like a spacecraft, streaking across the universe approaching the speed of light, impacting matter and virtual particles that is in its way, maybe slowing it down – the faster it goes, the more matter in the interstellar medium it encounters, keeping it from obtaining “c”. Does this have something to do with it?

ANSWER:
Nothing "happens" to the mass; if you were riding on it, you would see no change. That is one reason why I prefer to not think of mass getting bigger, I prefer to think of momentum getting redefined. But, if you want to think about mass increasing, as interpreted by someone with respect to whom the mass is moving, here is how you can think about it: assuming that you accept that no mass can go faster than light, as you approach c it gets harder and harder to push it faster and you interpret this as the mass acquiring more inertia, another word for mass. So, at low speeds, a 1 N force on a 1 kg mass would result in an acceleration of 1 m/s2, but at a very high speed it might require 106 N to achieve the same acceleration which you would interpret as an increase of mass. Your ideas of trying to mechanically make sense of the mass increase, while creative, are off the mark totally.


QUESTION:
You know how The particles that make up a rock are constantly in motion. However , a rock does not vidibly vibrate. Why do you think this is .

ANSWER:
The amplitude of vibration of each atom is on the order of the spacing between atoms, 10-10 m, way smaller than you would expect to see.


QUESTION:
(Going straight up, like a rocket launch) If I impart a force of 89kN to a 6818kg mass and determine it will accelerate at 13.05 m/s^2, how fast does that mean it will accelerate over time (what is its average acceleration) assuming the same force stays applied to it, and what speed will it ultimately reach assuming a standard atmosphere?

ANSWER:
First of all, you have it wrong. The forces on the rocket are 8.9x104 N up and its own weight 6818x9.8=6.68x104 N down. So, the net force is 2.22x104 N up. So the acceleration is 3.25 m/s2 up. But, this is at launch, and you do not have enough information to know anything later. First, the rocket is presumably burning fuel and so its mass is changing so its acceleration will be increasing. Second, when it starts moving fast air drag starts to become important (usually proportional to the square of the speed of the object) which is an additional force down. Third, if it goes high enough the density of the air gets less which reduces the air drag at higher altitudes.


QUESTION:
if i have a 5 kg bird in a contained glass ball and when the bird is setting the total weight is 10 kg what is the weight if the bird flyes inside the contained glass ball does it measure 5kg or still measures 10kg best regards !

ANSWER:
I have answered this and similar questions before.


QUESTION:
How does the weight of an object influence the angle at which it begins to slide down an inclined surface? I can find lots of info on how weight effects acceleration but not the angle needed to start to move down a slope?

ANSWER:
What allows an object to sit at rest on an incline is static friction. However, there is a maximum amount of static friction which you can get: the most static friction you can get is fmax=
μsN where N is the force pushing the surfaces together and μs is called the coefficient of static friction and is determined by the surfaces. This is only an approximation, not some law of physics, but it works pretty well for everyday situations. When on an incline, N=mgcosθ and the force trying to push the object down the plane is F=mgsinθ; here θ is the angle of the incline, m is the mass, and g is the acceleration due to gravity. If it is not sliding, the frictional force is just equal to F. But as the angle is increased, eventually the friction is no longer able to hold the object, that is F=fmax, mgsinθmaxsmgcosθmax, so tanθmaxs. So, you see, the answer is that the weight has no influence.


QUESTION:
This may be a random question and It might be too obvious, but I was wondering whether gravity violates the law of conservation of energy. Gravity seems to me, a constant force that is constantly acting on all objects but has no input energy to contribute to it.

ANSWER:
So, consider a mass, say the earth. It has a gravitational field which has some energy density. Similarly, it has mass so it has mass energy. That takes care of all the energy associated with that mass. Now, you can say the same things for some other nearby mass. If that nearby mass is a baseball, we have found it useful to introduce a potential energy function. But I always tell my students that this may be thought of as a bookkeeping device which automatically keeps track of the work which the earth's gravitational field does on that baseball. Before we introduce potential energy in elementary physics, all we have is that the change in kinetic energy equals the work done by all forces external to that object. So, before we "invent" potential energy, the work done by the baseball's own weight causes its energy to change. We do not worry about nonconservation of energy here because energy conservation only applies to isolated systems, systems which have no external forces on them, and the earth is most certainly an external force on the baseball. Adding potential energy to the system simply "internalizes" the field, it is no longer considered an external force. Now, you are worried about where the earth gets the energy to give to the ball. The answer is all in relativity. Suppose that you take a ball and move it out to infinity from the earth's surface. Do you do work? Of course! But, what happens to the energy you put into that ball-earth system? It shows up as mass! You know, E=mc2. This kind of argument is made all the time in nuclear physics where the mass changes are really obvious because the forces are so much stronger than gravity, but the same argument can be made for any attractive force. This mass change is where energy of chemistry comes from
e.g. when you burn coal you add an O2 to a C and get CO2 and a little less mass. When you split a U nucleus you lose a really significant amount of mass. A ball on the ground has less mass than a ball halfway to the moon. Strange, but true!


QUESTION:
why does a swimming pool that is 30 degrees celcius have more thermal energy than a cup of soup that has 60 degrees celcius

ANSWER:
The energy content of a cup of water is proportional to the absolute temperature. So a cup of water at 600C=332 K has energy Ehot and a cup of water at 300C=302 K has energy Ecold; so, Ecold/Ehot=302/332=0.91, the cold water has 9% less energy than the hot water. However, if you have two cups of cold water, the cold water has twice the energy of a single cup and so now Ecold/Ehot=604/332=1.82 so the cold water has 82% more energy than the hot water. And, so on

BETTER ANSWER: (Thanks to Michael Weissman at Ask the Van)
"
This sort of comparison of thermal Es assumes you have some defined zero, presumably E at T=0. (We needn't worry yet about at what p or V.) So thermal E(T) is whatever had to be added from there to reach T. That isn't even remotely close to proportional to T. These aren't ideal monatomic gases far above the boiling point. It's largely dominated by the latent heat of melting, an almost constant energy density throughout the liquid range." This clearly shows that my quantitative comparison was not correct; be assured, however, that the the soup has much less energy than the pool.


QUESTION:
A few days ago my friend & I were debating about how we could find the force with which a bullet of mass m and UNIFORM velocity v would hit a wall x m away. I said that the total energy would at the moment of impact would be equal to the kinetic energy of the bullet. So the total work done would be equal to the kinetic energy right at the moment of impact. So we could easily determine the kinetic energy by dividing the product of the mass and velocity by two. Then the result would be equal to the work done by the bullet. We could find the force by the formula W=fx. Please tell me if I am right.

ANSWER:
You are not right. You are right that the kinetic energy of the bullet is lost. However, x which you define as the distance from the wall to the gun, has nothing to do with it. Work is the force times the distance over which it acts and the force is not acting all the way across the room. What you might say is that if the bullet penetrates a distance x into the wall then the average force may be written as F=(
½mv2)/x. But, that assumes that all the work done by the force results in stopping the bullet, but some, much maybe, of that kinetic energy is converted into heat and sound energy. (Incidentally, uniform velocity has nothing to do with it, all that matters is how fast it is going when it hits the wall.)


QUESTION:
Are sound waves simple harmonic motion? Then are all waves simple harmonic motion? Which are and which aren't? How could you know the difference?

ANSWER:
Simple harmonic motion is defined to be a pure sinusoidal motion. That is, whatever (x) is vibrating must be expressible as x=sin(
ωt). Only pure, single frequency sound waves are of this form. So almost no sound is simple harmonic. All sounds, however, may be expressed as sums of simple harmonic sounds of different frequencies; this is called Fourier analysis of a wave.


QUESTION:
In strongman contests, the contestants pick up a telephone pole at one end, push it upright and flip it over. The other end of the telephone pole stays planted on the ground. Typically the telephone poles weigh 300 lbs and are 30 feet long.

  1. How would you calculate the actual weight of the pole being flipped at different points during the lift? For example, in the course of one flip we could say the pole travels in a 180 degree arc. The person is physically lifting for the first 90 degrees or less when gravity takes over. When the person picks it up off the ground it might be at 5 degrees, once it is over their head it could be at 30 degrees, once they take 2 steps forward it may be at 45 degrees, etc. - how do you calculate the actual weight of the pole along that arc at different points?

  2. As the person walks forward to flip the pole over their arms remain over their head (i.e. in this example once they're at or past the 45 degree mark in the arc), but it seems the weight would lessen along a curve as the pole gets higher and gravity begins to assist? That is - the person (fulcrum?) might be moving in - say - 3 ft. increments as they walk forward to gain leverage to flip the pole over. How is the weight being lifted adjusted to account for that variable?

ANSWER:
First, some terminiology: the weight of the pole is 300 lb, that is its actual weight. What you want, I believe, is the force you need to exert to hold it at any given angle. This is a problem involving torques
—you say that, in order to hold it at some angle, the torque you exert is equal to the torque the weight exerts, a bit more to move it. (Torques are calculated here about the point where the pole touches the ground.) The torque due to the weight is easy, simply 300x15xcosθ=4500cosθ ft-lb where θ is the angle the pole makes with the ground. The torque you exert depends on where you push and the direction you push, an infinity of possibilities. I have calculated for the two extreme possibilities: you push vertically up or you push horizontally. For the vertical pushes I assume that your hands are 7' above the ground; for the horizontal pushes I assume that your hands are 5' above the ground. The graph above shows vertical and horizontal forces you must exert vs. angle. Clearly, you want to start your lift by exerting a vertical, not horizontal, force. When you first lift it, you need to exert an upward force of 150 lb and when you get to 13.50 your hands are at 7' and you can start moving forward. When you get to a little more than 500, it starts to become more advantageous to push horizontally, becoming much more advantageous as you near the end. If you go real slowly, you will have to exert a maximum force of about 500 lb, but when the force you need to exert is small at the beginning you should push harder, thereby giving it some added speed which will help it over the middle angles where the force is maybe bigger than you can exert. Also, you will, through practice, learn to shift smoothly from vertical to horizontal .


QUESTION:
My question is in the nuclear equation (proton + electron-> neutron). when this necular equations occurs, an x-ray is emitted. Why is an x-ray rmitted and not a phroton of a lower energy? Would the reason be?, because the photon is movind so fast it is observed as a form of light? or that gama phrotons have no mass or electrical charge, and they are pure electromagnetic energy. Or I'm I way off base?

ANSWER:
You must be thinking about electron capture, a beta decay process which competes with
β+ decay. Here the nucleus "grabs" one of the atomic electrons and combines it with a proton to make a neutron and a neutrino (you didn't include the neutrino). Now, think about it: there is a hole where an electon used to be in the atom, usually the K shell. An outer electron falls into that hole and makes an x-ray which follows the electron capture.


QUESTION:
is central force and centripetal force the same?

ANSWER:
No. A central force refers to a force which depends only on how far it is from some force center. For example, to a good approximation, the earth's gravitational force is a central force. Centripetal force is the force which is responsible for a centripetal acceleration and may or may not be a central force.


QUESTION:
I understand according to einstiens thought experiment - if you travel near light speeds from earth and come back - you would be thousands of years in the future. So as the subatomic particles move at near light speed are they not affected by this relativity theory - I mean are they moving in to future.

ANSWER:
Indeed, clocks on moving subatomic particles do run slower by the usual factor γ=1/√[1-(v/c)2]. Particles in an accelerator, usually protons or electrons, will not change over time, so you would not notice it since they do not carry little clocks along with them. However, if you make an unstable particle like a pi meson (pion) which has a half life of about 2x10-8 s when it is at rest, will last much longer if it is moving at a high speed. For example, if v/c=0.99, 99% the speed of light, then you would expect it to go a distance 0.99x3x108x2x10-8≈6 m if its clock ran the same as yours. But, when you observe such speedy pions, they will go much farther because their internal clocks are slow by a factor γ=1/√[1-(0.99)2]=7.1; they will go, on average, about 42 m.


QUESTION:
My question is regarding an elevator in free fall. I had a debate with my uncle about what would happen to a person in a free falling elevator. I asked him if the person would float. My uncle said no. He pointed out that a ball would not float if put on the floor of the elevator and the cable was cut, so why would a person? To see what would happen to a ball in a free falling elevator, I put a ball on a folder (the folder representing the floor of an elevator), and I dropped the folder with the ball. The ball stayed on top of the folder as they both fell to the ground. This indicated that what my uncle said about a ball staying on the floor of a free falling elevator was likely correct. However, when I researched the subject on the internet, I came across many articles that said that a person would float in a free falling elevator. So I'm a bit confused. Why would a person float but a ball wouldn't? So my question is: 1) Would a person stand on the floor of a free falling elevator or would he float above the floor in a free falling elevator (i.e. weightlessness)? 2) If the person would stand on the floor of a free falling elevator, please explain why. If the person would float above the floor, please explain why.

ANSWER:
First, let's ignore the effect of air friction or any other kind of friction on the falling elevator. You and the elevator are both in free fall with the same acceleration. You would feel weightless and you would have no motion relative to the elevator which you did not have before it started dropping. If you were standing on the floor, you would remain standing on the floor but you would not feel the floor pushing up on you like you normally would, so it is perhaps a misnomer to say you are standing on the floor
you and the floor happen to be at rest relative to each other and touching. If you were holding a ball and released it, it would float right where you released it. If you gave yourself a tiny push up with your toes, you would slowly drift toward the ceiling.

Now let's talk about the more realistic case of including air friction. The elevator would feel a drag force (like a parachute) which got bigger the faster it went and eventually would fall with a constant velocity called the terminal velocity. You, however, are at rest relative to the air you are in, so you would at first feel weightless but feel weightier and weightier and then, when terminal velocity was reached, feel normal, all the while standing on the floor. But, I don't think the second scenario is what you were thinking about because it is not what would be called free fall.


QUESTION:
what would be the apparent speed of light,emmited vertically from a train travelling at a velocity v,to an observer who is standing in the platform

ANSWER:
The speed of light is always the same to all observers, so the answer is c=3x108 m/s. I suspect you are trying to use a trick to make the light go faster than the speed of light. You will not observe the light to go vertically but rather than at an angle forward in the direction of the train. When you apply the correct velocity transformatons, the vertical component will get smaller than c in just the right way that the vector sum of the horizontal and vertical components will be exactly c.


QUESTION:
My musical instrument has been broken during an air travel, but its container, a hard case with thick foam padding inside and outside is absolutely intact! how is this possible ?

ANSWER:
Suppose that it is dropped from a height of 3 m, about 10 ft. It would hit the floor with a speed of about 8 m/s. Suppose the thickness of the foam is 5 cm, about 2 in. Then the instrument would stop in a time of about 0.013 s and would experience an acceleration of about 640 m/s2 which is about 64g where g is the acceleration due to gravity. What this means is that, to stop your sitar in this short distance would require an force of about 64 times bigger than its own weight. Even dropping from 1 m would result in an acceleration of about 20g. If the padding were well designed so that this force was distributed over the whole surface of the instrument, it might survive, but probably one part will take more than its share of the impact. So, the padding in your case should be thought of as protecting the instrument from normal bumping and jarring, not major drops.


QUESTION:
There are a few aircraft like the X-15 (A) that are released from other aircraft (B), which obviously then fly at a higher speed than the 'carrier' (A). In my mind, if another aircraft (C) could then be fired from B, then it would be quicker still. In theory, could C release D, D release E and so on until an insanely high speed is reached. If it were possible to build the system I have described, is there any limit to the speed it could reach? Could it not even exceed the speed of light?

ANSWER:
Your idea has already been invented and is called a multiple-stage rocket. Don't get any fancy ideas about exceeding the speed of light, though! Ain't gonna happen.


QUESTION:
I was watching a program about the universe and the narrator said something along the lines of "this is what a neutron star sounds like", and then went on to show a sound. How is this possible if sound is a compression wave, and therfore requires a medium of some sort? I thought that space was a vaccuum, and therefore sound waves can't travel through it so, how can we hear what a star sounds like?

ANSWER:
These stars do not make sound, they radiate electromagnetic waves in pulses with the frequency of the rotation. So, if you take these pulses, put them through an audio amplifier and feed them to a speaker, they make sound with the frequencies of the pulses. To hear some, look here.


QUESTION:
My girlfriend and I were wondering whether an (1) airplane travels faster over the ground as it flies east to west due to the earth's rotating underneath it, leaving out tailwind or headwind. She thinks it does not travel fast because it is in the atmosphere and she equates an example she recalled where if someone in a bus going 80 mph (2) throws a ball straight up in the air, the ball doesn't go backwards. That just sparked an argument which I'm losing. I said it does go backward, you just can't notice because it's not in the air long enough to see a difference. It seems to me that if the bus was tall enough and you could throw it straight up high enough, the ball would come down behind where it was thrown because it is slowing down while in the air. (3) After a bit of searching, it seems that the consensus is that the ball will come down in the "exact same" place due to what some call conservation, others preservation of momentum. I understand that the objects inside the bus are going 80 mph just as the bus is - and the bus is a container. I don't understand why the ball thrown up on the bus does not start to decelerate AT ALL after leaving the person's hand if it has not contact with anything else beside the air in the bus. It seems to me that it must decelerate some. I know it would not rapidly decelerate as it would if thrown out of the window into oncoming air; and as compared to a ball thrown 80 mph in a ballpark, I could see if the deceleration on the bus is less rapid because the air on the bus is going 80 and maybe it pushes the ball forward some. But it seems to me that the air on the bus shouldn't have much more effect than an 80 mph wind would have on a ball thrown up by a person standing outdoors. And maybe the plane does not get an increase in ground speed exactly equal to the speed of the earth's opposite rotation due to whatever reasons, but it seems it should get some increase. (4) The atmosphere is not as fixed to the earth as a tree, is it? Doc, (5) I'd really rather you say something that can be used against my girlfriend here, but any help will be appreciated. I've thoroughly enjoyed the site and now wish I had taken physics.

ANSWER:
Whoa! Too many things going on in this question
—airplanes, balls, buses, girlfriends…I have taken the liberty to try to organize your question by numbering parts of it. There are two issues here. First, you always have to specify a velocity with respect to something else. What someone on the ground measures and someone on the bus measures are different. Second, you have to decide whether air friction is important or not to discuss the velocity of something which moves through the air.

  1. The airplane flies relative to the air and if the air is at rest relative to the ground, it makes no difference which direction the plane flies, it travels the same ground distance in a given time.

  2. If the air in the bus is at rest relative to the bus and if the bus is traveling with constant speed in a straight line, the ball will go straight up and straight back down as seen by somebody on the bus. Somebody on the ground will see the ball travel in a parabolic path. These statements are if air friction is negligibly small.

  3. The reason the ball comes down to exactly the same place in the bus is determined by what forces act on it. If you neglect air friction, the only force is gravity which acts straight down causing it to slow down going up and speed up going down. If air friction is not negligible, it is a force which always acts in the direction opposite the velocity vector, that is it never has a component which is horizontal which is what you need to make it start moving (relative to the bus) horizontally. If air friction is important, like if you throw it really fast and really high, it will still move vertically but it will take different times to go up and down. Unless there is a horizontal force (a wind, for example), something moving in the vertical direction will move only in the vertical direction.

  4. Since you stipulated no head wind or tail wind effects, yes, the atmosphere is just as rigidly attached as a tree.

  5. Sorry, your girlfriend has a much better grasp on this than you have!

One final proviso: remember where I said "traveling with constant speed in a straight line"? Well, this is never rigorously true because the earth is rotating and therefore Newton's laws are only approximately true. The effects of this (centrifugal and coriolis forces) are so tiny that any ball throwing will be unmeasurably small. However, for very long-range problems they can be important. These "forces" are what are result in the circulation of weather systems.


QUESTION:
It is well known to us that the energy is conserved everywhere in nature, infact it is the basics law in physics.But how far it is correct practically? mean to say if suppose we consume 100j of energy(inform of our food),can the net output work be of 100j or less than that or more than that?

ANSWER:
When you say you "consume 100 J of energy", you are really talking qualitatively, not quantitatively. Your body is a machine with inefficiencies. Nevertheless, the law of conservation of energy is certainly true. It states that the total energy of an isolated system remains constant, so any energy in one form always ends up in another form. Biological systems are more difficult to do experiments with because the isolation bit is hard to achieve. There is one little proviso on energy conservation, but it is not really what you are asking about.


QUESTION:
Do black holes "consume" dark matter?

ANSWER:
If dark matter exists (see my reservations about this), then it interacts only via gravity and would therefore be "consumed" by black holes.


QUESTION:
I know scientists have created small amounts of antimatter,but is all antimatter created equally or are there different elements ie: anti-hydrogen , anti-carbon etc.?

ANSWER:
Antihydrogen has been experimentally created. It consists of an antiproton orbited by a positron (antielectron). To the best of my knowledge, no other antiatoms have been created but, in principle, could be. They would consist of nuclei with antiprotons and antineutrons orbited by positrons. Really hard to pull it off, though.


QUESTION:
How much Electro Magnetic Radiation (in milligauss) would come off of your typical 365 kV electrical power line? Knowing distance is a factor could you give me levels at 50, 100,150, and 200 yards?

ANSWER:
I think you want the magnetic field strength which is not the same thing as electromagnetic radiation; certainly, this is what a milligauss measures. It is not the voltage which determines the field, it is the current which is flowing. Since power is current times voltage, the bigger the voltage, the smaller the current (and therefore the magnetic field) for a given power transmission. Suppose the power is 1 megawatt. Then I figure the current would be around I
≈3 A. The magnetic field a distance R from a current I is given by B=μ0I/(2πR)=0.06/R (measured in gauss, R in meters). So, for R=100 m, B=0.6 milligauss; you can do other calculations if you want. For comparison, this is about 1000 times smaller than the earth's magnetic field, so if you are worried about health issues, don't.


QUESTION:
Theoretically, if you were traveling through space in an object e.g., a spaceship and going at an incredible speed in one direction what would happen to you, inside, if it made an immediate 90 degree change in direction? wouldyou smash against the inside of the craft?

ANSWER:
First of all, you cannot make "an immediate 90 degree change in direction". It would require an infinite force to stop your forward motion simultaneous with an infinite force to start your sideways motion. So, you have to say you make the turn through a curved path. As an example, I will choose "an incredible speed" to be 104 m/s
≈22,000 mph, just a little faster than the shuttle goes. Now, at that speed let's make a 900 turn around a curve of radius 1 km=103 m. Then the spaceship and all its contents experience a centripetal acceleration of ac=v2/R=(104)2/103=105 m/s2. This is ten thousand times the acceleration due to gravity which means that it would take a force of 10,000 times your weight to move you in this circle. I would hate to see you after this maneuver, certainly you would not be recognizable.


QUESTION:
can you help with a discussion about how it is possible for heavy aeroplanes to take off and fly. We are aware of two principles - the aerodynamics principle in which the wing (and sometimes the fuselage itself) provides the lift, and the old principle of "speed defies gravity". were early pioneers such as the wright bros. aware of BOTH principles, or was the "speed" principle used alone to begin with, then the aerodynamics principle discovered and used to improve? Also, do helicopters use both principles or the aerodynamic principle only? There must be other principles as well, for example when thrust alone is used for a vertical take off craft, but we are mainly interested in the use and "time scale use" of the two i have asked about.

ANSWER:
"Speed defies gravity"? What is that supposed to be? I have discussed how an airplane works in some detail in an older answer. There are two important factors in providing lift for an airplane, Bernoulli's principle (which I guess you call the aerodynamics principle) and Newton's third law as you will see in the earlier answer. When you refer to speed defying gravity, are you suggesting that the thrust of the engines lifts the airplane? This thrust is much less than the weight of the airplane (trying to lift the airplane off vertically would be futile) so it is a small factor in lifting the weight (unlike a big rocket which lifts the load this way). The Wright brothers were the first to design and build wind tunnels and therefore experiementally designed the shape of the wings to provide lift. Helicopters are much like airplanes except they move the wings through the air rather than move the air over the wings. That is why airplanes are often called fixed-wing aircraft; helicopter rotors are actually wings.


QUESTION:
I want to ask about weight shift in vehicle. Weight shift is a very common terminology among car enthusiast (but most of them not physicist or engineer, some might not even study science well). Magazine always comment an accessory will help in weight shift of the car, etc. The common qualitative statements made by them in the use of weight shift are: (i) When a car turn, the weight shifts to outside wheels more; (ii) When a car accelerates, the weight shifts to the rear; (iii) when a car brakes, the weight shifts to the front. The above is just some statements to clarify what I meant by the term "weight shift".

Here is my question: Is "weight shift" a misnomer, or an ill concept of physics (vehicle dynamics to be specify)? The weight of an object is determined by the gravitational pull, regardless of what motion the object is doing, so how can the weight of different sections of the car change during maneuver? I think if one was to say force acting on different sections of the car during maneuver is different, then it is correct; but to say the weight shifted during maneuver, I can't get it.

ANSWER:
Technically, weight shift is a misnomer, but it is not too bad because it is a shift in apparent weight. Normally, we think of weight as acting at the center of gravity of something, but if that something is accelerating, it may appear to move elsewhere. I will try to explain. In all three of your examples, the qualitative observations are essentially the same. For the accelerating car, the "apparent weight" shifts toward the rear wheels, opposite the direction of the acceleration. For the braking car, it is just the same as the accelerating car except that the direction of the acceleration vector is toward the back, the "apparent weight" shifts toward the front wheels, again opposite the direction of the acceleration. For the car turning there is an acceleration also, this time toward the center of the circle (called the centripetal acceleration) and here the "apparent weight" shifts toward the outside wheels, again opposite the direction of the acceleration. So, if I understand one of these I will understand them all. In every case, the force which causes the acceleration is the static friction between the tires and the road. The figure to the right shows a generic case. The center of gravity is at the point marked x. The forces labeled are the weight W, the force of the road up on the left tire N1, the force of the road up on the right tire N2, and the static friction force of the road on the tire f. If the car is at rest or moving with constant speed, f=0, W=N1+N2, and N1=N2; this assumes the center of gravity is halfway between the wheels, but if the center of gravity were moved to the left, we would have found that N1>N2. The three scenarios you mention can all be represented by this figure. If the car is accelerating, it is moving to the right and speeding up and the left wheel is the rear wheel; here f=ma where m is the mass of the car and a is its acceleration. If the car is braking, it is moving to the left and slowing down and the left wheel is the front wheel; again, it is the friction which causes the car to accelerate (slow down). If the car is rounding a curve, the center of the curve is to the right, and the left wheel is the outside wheel. We still have W=N1+N2, but now the two forces N1 and N2 are not equal. Since the car is not rotating about its center of gravity, if we sum the torques about x it must sum to zero:(N1L/2)-(N2L/2)-fh=0, so N1=N2+2fh/L. So, N1>N2, it appears as if the weight (center of gravity) has shifted toward the left. So, weight shift is a pretty good name for it, I would say; apparent weight shift would have sounded a bit pedantic for car buffs, don't you think? (By the way, I drew the friction only on the left wheel but if there were friction on the right wheel instead or in addition, the conclusions would all be the same because the torques from the right wheel friction would have been the same sign as the left.)


QUESTION:
i know what mass is in equations, and i could define it that way but in actuality what is mass?

ANSWER:
There are two ways you can think about mass. One is inertial mass, the property an object has which causes it to resist being accelerated; the more mass something has, the harder it is to accelerate. The other is gravitational mass, the property an object has which allows it to feel or cause gravitational forces. It turns out that they are the same thing, which is sort of the basis of the theory of general relativity, our best theory of gravity to date. If you want to know why objects have mass, that is a different issue; it is thought that the thing which is responsible for endowing particles with mass is called the Higgs boson, currently being anxiously searched for at particle accelerators.


QUESTION:
While reading " A brief history of time by stephen Hawkings, i found that he explained something about spin 0,1,2. I wanna know that how is spin 0,1,2 different from Spin 1/2 and how does pauli exclusion principle explains it?

ANSWER:
All these numbers are spin quantum numbers. If the spin quantum number of a particle is s, then the intrinsic angular momentum of that particle is S=
{√[s(s+1)]}h/(2π). Intrinsic means that this is not due to some orbital motion, but is associated with even a free particle. It is sort of like the earth whose orbital motion around the sun and rotational motion about its axis are different things; particle spin is sort of like the earth's rotational motion about its axis. Particles with half odd-integer spin quantum numbers (1/2, 3/2, etc.) are called fermions and obey the Pauli exclusion principle; examples are electrons, protons, and neutrons (all with s=½). Particles with integer spin quantum numbers are called bosons and do not obey the Pauli exclusion principle; examples are the pi meson (s=0), photon (s=1), and the graviton (if it exists, expected to have s=2).


QUESTION:
Assuming an object is in motion and traveling through space under its own inertia, let’s assume you stop it. In order to stop it you apply a force which results in a negative acceleration of the object causing it to stop. But since a force is met by an equal and opposite force, what is the force of the moving object?

ANSWER:
Newton's third law says that if you exert a force on the object, the object exerts an equal and opposite force on you. But that force is on you, not the object, and so it has no effect on the object.


QUESTION:
I have a shipping container that weighs 8000lbs. I am needing to lift it up to do some digging underneath it (to level it out). I have a 2 1/4 ton jack (4500 lbs) that I am wanting to use to lift the container from one end at a time. My question is how much weight am I actually lifting (from the one end) as opposed to the full 8000 lbs?

ANSWER:
The crucial consideration is if the weight is uniformly distributed in the container. In other words, could you balance it on a point in the middle? If the answer is yes, you should be ok as long as your jack lifts straight up vertically. It will hold about 4000 lb. If the weight is not uniformly distributed, then when you lift the end where more of the weight is, you may exceed 4500 lb.


QUESTION:
I have often heard of Newton's first law being referred to as the "law of inertia" but wouldn't it be much more appropriate to call the second law the "law of inertia?" The fact that a body with no forces acting on it maintains a constant velocity really just seems to establish the concept of an inertial reference frame and is true for objects of any mass. The concept of inertia seems to be represented by the second law by describing the resistance of matter to acceleration.

ANSWER:
Think about what inertia means in everyday life. If we say I have a lot of inertia, it means that I am unlikely to change my mind if not pushed. Anyhow, it is just semantics, so why worry about it? I agree with you that one interpretation is that the first law is important in that it defines an inertial frame of reference, not as a special case of the second.


QUESTION:
How do I know when a scenario is an example of Newton's first or second law? For example, when a bicycle loses speed as it coasts uphill. Is that straight line motion but the other forces acting on the bike are friction and gravity, so it doesn't remain in straight line motion? Or is it that the net force, including friction and gravity, results in acceleration in the same direction?

ANSWER:
Newton's first law is appropriate for an object moving in a straight line with constant speed. The bike losing speed is moving in a straight line, but not with constant speed, so Newton's second law is appropriate. Newton's first law means that if the velocity (magnitude and direction) is constant, the sum of all forces on the object is zero. If the velocity is changing (acceleration), the sum of all forces (taken vectorially) equals the mass times the acceleration.


QUESTION:
I am using "Shockwatch" indicators to measure the impact of baseballs on a helmet. I am comparing the impact of a baseball with and without a helmet. These are impact indcators from 10G to 100G. How do I make the data meaningful converting the G's to _________. Would it be newtons of force? I don't know. I am a 7th grade life science teacher helping one of my students.

ANSWER:
The "shockwatch" is likely just an accelerometer which measures average acceleration of something over the impact time. The "G" refers to the acceleration of gravity which is 9.8 m/s2=32 ft/s2; this means that a freely falling object speeds up by 9.8 m/s as each second ticks by. So, if the baseball stops in 0.01 s and is initally going with a speed of 60 mph=27 m/s, the average acceleration is 27/0.01=2700 m/s2=2700/9.8=275.5 G. To convert this to a force, you need to know the mass of the object. The mass of a baseball is about 0.14 kg, so the force over the impact time is F=ma=0.14x2700=378 N=85 lb. You probably have your shockwatch attached to the object you are hitting, so you need to know its mass if you want to know the force on it. So, if you measure an "impact" of 50 G, and the mass of the head is 4.5 kg (10 lb), then the average force is 4.5x50x9.8=2200 N=495 lb.

Important conversions: 1 N (Newton)=0.225 lb (pound), 1 kg (kilogram) weighs 2.2 lb


QUESTION:
If you took a waterproof bathroom scale that was tared correctly to register zero on land and dropped it into 30m water, would the weight of the water register on the scale?

ANSWER:
A scale is essentially a spring which indicates weight by the amount of compression. So, think of a spring in a flexible box. If you took it down to some depth there would forces on all sides. But, the top force would be down and equal to the weight of water above it and the force on the bottom would be up and equal to the weight of the water above it. So the spring would compress just as if a column of water of cross section equal the the area of the scale were being weighed. (I have neglected the thickness of the scale; the forces on the top and bottom are not quite equal being at slightly different depths.)


QUESTION:
at one point in Nova telecast "What is space?" Brian Greene explains that gravity can, as a result of Einstein's insights, be viewed as a consequence of spacetime geometry rather than as a force. Why then do physicists devote so much effort in treating gravity as a force; moreover, as force that must be accounted for in any successful "theory of everything"?

ANSWER:
I have previously answered this question.


QUESTION:
I was wondering how it is possible to see some of the first stars and galaxies in the young universe? I understand that we are seeing light from them because the light has taken billions of years to get here but I don't understand how the atoms that make up our bodies and our galaxy got here from ancient stars who's light is just now getting here. That would seem to suggest that matter traveled faster than light.

ANSWER:
The heavy atoms from which you are made were, indeed, manufactured inside stars long dead. However, those stars were not among the most distant in the universe, they were stars in our own neighborhood, our own Milky Way galaxy.


QUESTION:
Why is it difficult to calculate the terminal velocity for a cat falling from a high roof top?

ANSWER:
I do not know what you mean "difficult to calculate". We can estimate it pretty easily, but certainly not do it precisely. First of all, any calculation having to do with air friction is going to have approximations and assumptions. For something like a cat, roughly 2 kg (4.4 lb), falling, it is a very good approximation to say that the drag force is proportional to the square of the velocity. It turns out that a fairly good approximation for the force is F=
¼Av2 where A is the area the falling object presents to the onrushing wind and v is the velocity (this is only for SI units). Since it depends on A, it depends on how the cat orients itself: if in a ball he will fall much faster than if all spread out. Suppose we take the area of a falling cat to be about 20 cm x 40 cm=0.08 m2. Then the force will be about 0.02v2. Now, the cat's weight is about mg=2x9.8≈20 N. When the force of air friction is equal to the weight force down, the cat will fall with a constant velocity called the terminal velocity: 0.02vt2=20, so vt=√(20/0.02)≈30 m/s=67 mph. If you google "terminal velocity of a cat" you will find the number 60 mph, so my approximations were evidently reasonable. There, now, that wasn't so difficult, was it?

RELATED QUESTION:
I was asked what the terminal velocity of an unladen sparrow is. I read that the average weight of a field sparrow is .5 ounces. Approximately 5 inches in size with a wingspan of 7.9inches. Lets say falling from 50ft?

ANSWER:
Look at the answer above. If the sparrow falls straight down with wings stretched out, I would estimate his area to be
½x5x7.9=20 in2=0.013 m2. Following the same as for the cat, but using m=0.5 oz=0.014 kg, I find vt≈6.6 m/s=15 mph=22 ft/s.


QUESTION:
Hello, could you tell me about the relative speed of a plane traveling let's say 400 mph. How fast would it appear to be moving from the ground at different altitudes 3000 ft compared to 30000 ft.

ANSWER:
How fast something "appears to be" is meaningful only in comparing it with something else. To compare apparent speed at 3000 and 30,000 ft I would compare their angular velocities, that is how long does it take to traverse some angle, in particular compare the time for each to move from one point in the sky to another. The distance the higher plane would have to travel to have the same angular displacement is ten times as far, so it would appear to be going ten times slower.


QUESTION:
According to the molecular biology book I am reading, when a chemical bond breaks, the energy in the bond is released in the form of heat. which is to say, when a chemical bond breaks the entire atom's speed increases (the atom gains 'heat energy'). But what is happening at the electron/proton level to make this happen? I know that when some bonds break they are 'exothermic' (give off heat-- ie,make the atoms breaking apart move faster) and when other bonds break they are 'endothermic' (absorb heat-- ie, make the atoms breaking apart move slower) but what is actually physically happening?

ANSWER:
You are talking about chemistry here, not physics, but chemistry is really just atomic/molecular physics so I will comment. You have some misconceptions here which need to get cleared up. Suppose you have a molecule, say O2. These two oxygen atoms are bound together and I guess a chemist says there is a "bond" between them. If you simply break this bond, you will never get energy (exothermic), you will always have to add energy (endothermic). If you did not have to add energy to make O2 be O+O, why would the molecule stay together on its own? This is true for any stable molecule
—it is going to cost you energy to break it apart. You need to create a bond, not break one, to get energy out. So, if you start out with O2 and are able to cause a carbon atom to attach to it, you get energy released: O2+C—>CO2+energy. You have discovered fire! More complicated molecules might have the property that if you break a bond you make the others stronger and so that could release energy; it will always cost energy to break a bond, but if the bonds in the new molecule have increased by more than your cost, there is a net gain.


QUESTION:
If you give a object at rest a push up a ramp, does it take longer, the same, or less time to slide back down to the same position? Assuming there is sliding friction. I would think the same due to conservation, but then I started thinking about the influence of gravity and friction and am now, I'm just confused.

ANSWER:
Energy is not conserved here because the friction is taking energy away. (Actually, that energy ends up in heat, which is why the object and ramp will be slightly warmer afterward.) To answer your question, think about forces. There are two forces which affect the motion along the ramp, the frictional force which is always opposite the direction the object is moving, and the component along the ramp of the object's weight which is always down the incline. So, on the way up these two forces both point down the incline and on the way down the weight points down and the friction points up. Therefore the net force slowing the object down when going up is greater than the net force which is speeding the object up when it is going down. Therefore the time going down is longer than the time going up. Of course, if the friction is big enough it well never go back down and, certainly, infinite time is longer than however long it took to go up.


QUESTION:
Who are one of the five con-men who believe or proclaim they have found ways around the law of conservation of energy? Just would like to know one of them and what they believed to work. Thanks physicist!

ANSWER:
What are you talking about?

FOLLOWUP QUESTION:
Was it aristotle , socrates, maxwell, or someone else who believed that they found a way around the law of conservation of energy ?

ANSWER:
Well, the notion of energy was not proposed until long after Aristotle or Socrates were around, so we can rule them out. I never heard of Maxwell thinking energy conservation might not be true. Niels Bohr proposed that perhaps energy was not conserved in nuclear beta decay, but then the neutrino was proposed by Wolfgang Pauli, and so energy conservation was saved. I guess we should state what energy conservation is: the total energy of a system remains constant as long as no external forces do work; or, put more simply, the total energy of an isolated system never changes. However, the Heisenberg uncertainty principle allows for energy to be not conserved as long as it is only for a very short time; the total energy E of an isolated system may change by an amount
ΔE but for no longer than a time Δt such that ΔEΔt<h/2π, where h is Planck's constant, h=6.6×10-34 m2kg/s.


QUESTION:
Galaxies recede from us at a velocity proportional to their distance. Why then is Andromeda on a collision course with our beloved Milky Way?

ANSWER:
The overall average behavior is that the universe is expanding. However, at distances much smaller than the size of the universe there is no fundmental reason why everything has to be moving apart. For example, the sun is not moving away from us. And, there are attractive gravitational forces which can be important between nearby galaxies. Astronomers have observed many examples of galaxies colliding. The Milky Way and Andromeda galaxies will collide in about 5 billion years. You can see a simulation of that collision at this link.

 

 

 

 


QUESTION:
I understand that the power that a stream of moving water can generate or develop is proportional to the velocity of the water current, cubed. Can you explain where this comes from?

ANSWER:
I never saw this before, but I think I can figure it out. The power P generated by a force F may be written as P=Fv where v is the speed of the agent applying the force (the water here). Now, if an object moves through a fluid, it experiences a drag force which, for most cases of interest, is proportional to the square of the velocity, F=Cv2. If the fluid exerts that force on the object moving through it, then if it is the fluid that is moving instead of the object, the force must be the same. Therefore, P=Cv3.


QUESTION:
what is the upward force in pounds on the bottom of an empty basin caused by a groundwater depth of 8 feet above the tank bottom? The basin has a base of 300 square feet. What is the formula? This is in regards to not emptying a tank fully for repair as the upward force may displace it from its resting place.

ANSWER:
The key here is to find the pressure at a depth d (8 ft in your case) below the surface of a liquid of density
ρ. This is given by P=ρgd where g is the acceleration due to gravity; when you know the pressure, you can find the force F on the bottom by multiplying the pressure times the area A (300 ft2 in your case), F=PA=ρgdA . So, there is your formula. You are probably wanting to do this in English units, but physicists hate to work in English units. I did it in SI units where ρ=103 kg/m3, g=9.8 m/s2, d=8 ft=2.44 m; so I found P=2.39x104 N/m2 and converted to P=499 lb/ft2. So, F=499x300=149,700 lb. Actually, I found that it is pretty easy to do this in English units: the weight density of water is ρg=62.4 lb/ft3 and so F=62.4x8x300=149,760 lb.


QUESTION:
If the LHC accelerates particles almost to the speed of light, why don't they get huge and approach infinite mass?

ANSWER:
They do approach infinite mass but really have a long way to go yet. See an earlier answer.


QUESTION:
Lets say its 25,000 light years to the center of our galaxy. Meaning, light taking off from earth today would take 25,000 years to get to the center of the galaxy, or vice versa, from the center of the galaxy to us. Let's say we had a ship that could travel at 99% the speed of light, and we board that ship and leave earth travelling at 99% light speed. From earth, as we watched the ship travel, we would see that it took almost 25000 years to reach its destination. But what about the perspective of those on the ship? How long would the journey seem to take for the travellers, moving at 99% the speed of light? How much time would have elapsed by their watches?

ANSWER:
The important factor in relativity is
γ=1/√[1-(v/c)2]. In your case, γ=1/√[1-(0.99)2]=7.1; so the traveler sees the distance d he has to travel to be shortened, d'=25,000/γ=3520 ly. So, the time it takes is t'=d'/v=3520/0.99≈3560 yr. You could also get this from the time dilation formula, t'=t/γ=(25,000/0.99)/γ=3560.


QUESTION:
Hey dude, With the endless amount of stuff floating around in space, how come none of it hits earth? And to be more specific, how come nothing so massive and unstoppable by earths defensive capabilities ever comes towards earth and destroys it?

ANSWER:
Hey Dude back at ya! It happens all the time
—what do you think shooting stars are? And, tons of space dust hit the earth every year. In the early solar system, there were a lot more pieces of stuff "floating around" and there were many more collisions with bigger objects than there are today. One collision with an asteroid-sized object is thought to have caused a major extinction of life millions of years ago. But most remaining objects are in reasonbly stable orbits now. And, you just have no idea how mostly empty space is—just statistically it is immensely improbable that collisions will occur. Imagine two guys with rifles 5 miles apart firing their rifles; what is the likelihood that two of their bullets will collide? There was one major collision in 1994 when Comet Shoemaker-Levy hit Jupiter. Many asteroids have highly variable orbits which could possibly result in a collision with earth, but it is a very low probability event.


QUESTION:
If you take the same exact object and drop it from different heights like 1,000 ft... 5,000 feet... 10,000 feet... and 100,000 feet, will it hit the ground harder from a higher height or does it not matter how high you are once the object reaches terminal velocity. My husband says if you drop a rock from 10,000 feet it will do more damage than if you dropped that same rock from 3,000 feet. Is that true?

ANSWER:
If it reaches the terminal velocity before it reaches the ground, then it makes no difference what height it was dropped from. Just as a rough example, consider an object with a cross sectional area of about 0.04 m2 (like a 9" diameter ball) and a mass of 1 kg (about 2 lb). It would have a terminal speed of about 30 m/s (about 70 mph) and a characteristic time of about
τ=3 s. What the characteristic time means is that if the rock falls for a time much greater than the characteristic time, it will have achieved terminal velocity; e.g., if it falls for 3τ=9 s it will have achieved 99.5% of the terminal velocity. From 3000 ft it would fall about 14 s if there were no air and from 10,000 ft it would fall about 25 s. So both will certainly fall much longer than the characteristic time and therefore both would hit the ground with a speed of 30 m/s and have exactly the same effect.


QUESTION:
Please, can you explain me the process of how an electron can get accelerated with a magnetic field (like in a particle accelerator like CERN, by instance) ? In particular, from where do electrons get the increase of momentum: is the absorption of photons involved (since the EM fields work on the interchange of photons)?

ANSWER:
The accelerator (LHC) at CERN does not accelerate electrons, it accelerates protons. There are many electron accelerators, though. But, here is an important point: a magnetic field cannot accelerate an electric charge. Accelerators always use electric fields to speed the charged particles up. The fact is that a magnetic field cannot do work because the force it exerts on a charged particle is always perpendicular to the particle's path so it can change the direction of the particle's velocity, but not its speed. The reason you are thinking that magnets are responsible is that there always so many magnets in an accelerator laboratory. These are used to steer the beams of charged particles, like in the LHC the many magnets keep the beam circulating in a big circle.


QUESTION:
How many Newtons are exerted when a 300 lb. man falls 3 ft.?

ANSWER:
Read faq page. This question has no answer.

QUESTION:
We are working to produce a safety harness and the strap material we are using has a maximum Newton rating - we were trying to get an idea of what Newton rating would be needed to support a 300 lb. man if he fell 3 ft. Being hunters (tree stand safety harness) - perhaps we are wording the question incorrectly. Can you clarify your response?

ANSWER:
What matters is how long it takes the falling guy to stop. The mass of a 300 lb guy is about 130 kg, the acceleration of gravity is 10 m/s2, and so the weight of the guy is about 1300 N. You need that strong a strap just to hang him there at rest. If he falls 3 ft (about 1 m) he will be going about 4.5 m/s. So, let's call F the force needed to stop him and t the time it takes him to stop; I reckon that F
≈130(10+(4.5/t)). For example, if he takes ¼ s to stop, F≈3600 N to stop him. The straps are probably pretty unstretchy, so your best bet would be to make the harness out of a stretchy material because, don't forget, the bigger F is the more it is going to hurt during the stop.


QUESTION:
is proton diffraction is possible?

ANSWER:
Any particle, including a proton, will be behave like a wave if you look for it and diffraction is possible for all waves. In fact, I can give you an example of proton diffraction which I myself measured. What is shown in the figure to the right is the differential cross section for 800 MeV protons elastically scattered from the nucleus 90Zr plotted as a function of the angle (in degrees) where the protons were observed. Differential cross section is, essentially, the probability that the proton (wave) will scatter (diffract) to some angle. This is very much like the diffraction pattern you would observe for visible light striking a sphere
—diffraction maxima and minima. In fact, you can qualitatively understand this diffraction pattern if you calculate the wavelength of the protons, λ=h/p≈10-15 m, and approximate the positions of maxima by the double slit relation nλ=dsinθ where we take d to be the diameter of the nucleus. Taking the two consecutive maxima at 80 and 130, λ=d(sin130-sin80)=0.08d=10-15. So, the diameter of the 90Zr nucleus would be about 12.5x10-15 m=12.5 fm. As a check, the diameter of a nucleus with atomic weight A is well approximated as d≈2x1.25xA1/3 fm, so for A=90, d≈11.2 fm, pretty good agreement for such a rough calculation. The experiment was done at the Los Alamos Meson Physics Facility (LAMPF).


QUESTION:
I had an argument with my roommate I was hoping to resolve. Basically, can a dart bounce off of the metal rims of a dart board past the point from where it was thrown, say 8 feet. My point was that if the force is strong enough (enough force to go 50 feet w/o obstructions) when it hit the board there would be 42 feet (I dont know if this is correct) of force left over to push back. This would be enough to push it back further than the 8 feet. He seemed to believe that due to the angle of the collision, it would bounce off and not make it. I gave an example that from 1 feet it would surely bounce back far enough, but he didn't seem to agree. Is there a physics law or example that can explain. Note: Dart bars are curved, so a dart could come down on an angle and hit the top curve of the metal bar and get a different angle. But we were also arguing that even if the wall was flat, it could still happen.

ANSWER:
A very light object (dart) can bounce off a very heavy object (board attached to wall) with a speed about equal to the speed it came in with if the collision is elastic. You may know that the maximum range for a projectile is achieved (neglecting air friction) if it starts out at a 450 angle relative to the horizontal. Suppose the dart is thrown with an angle of say 100, so that it comes back to the same level it was thrown from when it gets to the dart board. Then, if it was thrown at that same speed but at 450, it would clearly go farther (assuming the board and wall weren't there), right? So, if  it bounces elastically off the "bar" at an angle of 450 it would go back farther than it came in from.


QUESTION:
Can you speed up the orbital speed of a electron around the nucleus?

ANSWER:
Technically, an electron does not have an orbital speed. The Bohr model of the atom, the "planetary model" is only a rough approximation as to what is actually going on. The best way to think about the electrons is a cloud around the nucleus. What you can do, equivalent to speeding up the orbital speed, is add energy to an atom which then places the electron in a higher energy state. This can only be done in discrete steps, not in any old amount you might wish to add.


QUESTION:
We're having this physics discussion at work (in the medical field). One of my co-workers insists that if you took a pound of anything and spread it out evenly on a scale, it would weigh less than if you piled it up in the middle of the scale. He said that the density of the mass of whatever you piled on the scale, would increase the gravitational pull moreso than if the object being weighed was spread out evenly, like say mashed potatoes, for example. What do you say?

ANSWER:
The stock answer to this question is that weight is the force that the earth exerts on a mass and that force is independent of how the mass is distributed. But, that is not quite right and your question is clearly a "hair-splitting" sort of question, so I will explain. The gravitational force between two objects depends on how far apart they are
—the farther apart, the smaller the force. The weight of something is determined how far it "is" from the center of the earth. If you compare the mashed potatoes spread out with the mashed potatoes heaped up, those heaped up are, on average, farther away from the center of the earth by a few centimeters, and therefore weigh less, not more. But, don't get too excited about this—the difference would be about one millionth of 1%, less than you could ever hope to measure. You would get a much bigger difference if you weighed the same thing upstairs and downstairs in your home, again something you know from experience is, for all intents and purposes, the same. [Rereading the question, let me address the density argument of your colleague. It is true (again to a really small degree) that the weight of the potatoes on the top of the heap will press down on those on the bottom and make the potatoes slightly more dense. However, a scale does not measure density, it measures mass and therefore two objects with different densities but equal masses would weigh the same.]


QUESTION:
Why do electron microscopes get better pictures than regular light microscopes?

ANSWER:
The problem with optical microscopes is that you cannot look at anything which is comparable to or smaller than the wavelength of the light. Visible light has wavelengths of a few hundred nanometers, ~6x10-7 m, so what you are looking at has to be bigger than that, a few microns maybe. The reason is diffraction, light does not just travel in nice straight lines when obstructions or aperatures get to be on the order of the wavelength, e.g. light can be bent around a corner. If light does not go in nice straight lines, geometrical optics doesn't work. On the other hand, we know that particles like electrons can behave like waves and their wavelengths are inversely proportional to their momentum. So, if we make an electron go fast enough it will have a wavelength much less than light and therefore let us "see" much smaller things.


QUESTION:
A fly is hovering in a car. the car is going 45 MPH with the windows up. the car hits a solid object and suddenly stops. Does the fly hit the front window, the back window, or nothing at all?

ANSWER:
This is tricky. The stock answer would be that the fly would smash into the front windshield just like you would if you weren't wearing your seat belt. But, the fly is actually hovering with respect to the air, and so whatever happens to the air will be what happens to the fly. The air does not all smash into the windshield but more or less stays just where it is relative to the car. The reason for this is that air is a collection of molecules most of which are already going much faster relative to the car than the car is relative to the ground. An average air molecule is going about 1200 mph. Think of half of them going (relative to the road) toward the back with speed like 1155 mph and half going toward the front with speed like 1245 mph, so their average is still 1200. When the car stops, the fly sees no significant change and just continues hovering where he is. He has such a small mass that he does not have enough inertia to overcome the air drag he would be experiencing and hit the windshield.


QUESTION:
Has there ever been a hypothesis that dark matter and or dark energy are forms of gravity? Does anything in string theory allow for this?

ANSWER:
Read my FAQ on dark matter/energy. You will see that my own feeling is that so-called dark matter may simply be symptomatic of the possibility that we do not understand gravity as well as we think we do; I seldom see astrophysicists or cosmologists adopt this view, but it is an open possibility until somebody actually directly observes "dark matter particles". Dark energy can be "understood" in terms of general relativity; Google cosmological constant. String theory allows anything and predicts nothing.


QUESTION:
how does ac currrent move fromone place to another, if is goes back and front.

ANSWER:
It does not "move from one place to another". For that matter, dc current almost does not either. The electrons move so slowly that it might take a day for electrons to move from battery to light bulb. The important thing is that all the electrons in the wire are moving almost instantaneously when you close the switch, ac or dc.


QUESTION:
Picture a perfectly aligned row of pool balls all touching each other that was 200,000 miles long. If the first ball was struck with enough force at the exact center of the ball, would the movement of the last ball occur before a laser beam could move from the beginning to the end of the balls?

ANSWER:
It would occur much later than the arrival of the light. The disturbance down the row of billiard balls travels with about the speed of sound. And, an important law of physics is that nothing can travel faster than the speed of light. See FAQ page.


QUESTION:
The sun produces enough gravity to hold the earth in orbit. So why dont space shuttles get attrcted towards the sun?

ANSWER:
It does. But the gravitational force from the earth is much bigger because, although earth's mass is much smaller than the sun, the shuttle is much closer to the earth. So the motion of the shuttle is much more determined by the earth than the sun.


QUESTION:
What is the "power" of gravity? i.e. how much force per second does it apply? Power = F x V, = m x g x V, but for V = 0, what power is required to make an object hover? A practical question might be if I have an electromagnet, how much power does it need to hover 1kg of metal?

ANSWER:
Force per second is not power, energy per second is power. To levitate something you must exert a force up on it equal to its own weight. To levitate a 1 lb object you must exert, somehow, a 1 lb force up for as long as you want it to levitate. No energy is consumed because the force does no work. Of course, if you are using an electromagnet, you must use power, but that is a feature of the design of the electromagnet, that work essentially goes into heating the coils of the magnet. If you had an appropriate permanent magnet, it could levitate something forever with no expended energy.


QUESTION:
I am doing an experiment with dropping coffee filters. By changing the mass (adding more filters) and keeping the distance the same (4 meters) and recording the time it takes for the stack to hit the ground. By doing this I find the velocity. Wouldn't there be an asymptote at 9.8, for the velocity of the dropped object?

ANSWER:
Technically, you are not measuring the velocity, you are measuring the average velocity. But, since coffee filters have such small mass, you can probably assume that terminal velocity is reached almost immediately after dropping. What you are studying is the fall of objects for which air resistance is important. I do not know what you mean by the asymptote at 9.8, but it is certainly wrong because 9.8=g is the acceleration due to gravity, it is not even a velocity. What you want is to find out how velocity depends on mass. This will depend how the terminal velocity depends on mass. Now, the force of air drag will depend on two things, the geometrical size of the falling object (but all will be the same, so you cannot study that) and the speed the thing is moving. Usually this force can be parameterized by F=Cv2 where C is some constant determined by the fluid (air) and the geometry, but not the mass. So, the faster it falls the bigger this force is and when it gets as big as the weight of the object it stops accelerating and falls with a constant speed called the terminal velocity which you hope to be determining. So, drag=weight when mg=Cv2 where m is the mass and g=9.8 m/s2. Solving, v=
√(mg/C); so, if you make a graph of v vs.m you should get a straight line if indeed the force is a quadratic function of the speed. And, from the slope, you could determine the constant C. If you do not get a straight line from your data, try using F=Cv; then if you plot v vs. m you will find a straight line.


QUESTION:
If I could detonate a firework inside a contained vacuum void of any other mass objects (simulating the big bag) and watched (over time) would I see the particles eventually attract to each other? Would the results of this sort of experiment prove gravity exists in every particle ever created?

ANSWER:
No. The gravitational attraction is very, very small between the little pieces of your firecracker. So the speeds the pieces had when it blew up would have exceeded the escape velocities from their neighbors and they would continue moving apart forever. Escape velocity is the speed something must have to escape from the gravity of something else; e.g. the escape velocity from the surface of the earth is about 7 miles per second, but this is for a much stronger gravitational force. Just to give you another example where the mass involved is still much larger than your masses but much smaller than the mass of the earth, the escape velocity from the surface of a baseball is about 8 cm/hr.


QUESTION:
What is overpowering gravity and causing the outward acceleration of the universe?

ANSWER:
If I could answer that question, I would be next in line for the Nobel Prize in physics. This year's prize went to the astrophysicists who discovered this. The glib answer would have been dark energy, because that is what astrophysicists call what is responsible for the acceleration. However, this is just putting a name on something we do not understand.


QUESTION:
My question has to do with traction and the movement of a wheel(a wheel alone). Traction is essential for its movement both linear and circular. But if we throw a wheel forward it rolls some meters and then it stops(and falls). Which force is responsible for the decrease in its velocity? Cause if traction is parallel to the ground facing backwards then linear movement 's negative accelleration is explained but not angular negative accelleration. If traction is parallel to the ground facing forward then angular negative accelleration is explained but not linear. If traction is zero then which force decreases both velocities linear and angular?

ANSWER:
One of the reasons I love doing Ask the Physicist is because I often learn things I did not know or had never thought about. You would think that a guy who has been teaching introductory physics courses for nearly 50 years would find this question simple. But, indeed I was puzzled by it because, as I have found by thinking about it and talking to some friends, I wasn't thinking beyond the friction force (which questioner calls traction) being simply the only force in the horizontal direction and obviously stopping the forward motion after some distance. I never addressed the angular acceleration of the wheel before. This answer will be long-winded because that is what I do when I have learned something which pleases me! What frictional forces are important to understand the rolling of a wheel? Most introductory physics classes talk only about the contact forces of static friction and kinetic friction. Kinetic friction is not applicable to this problem because the wheel is not slipping on the ground, and static friction might be important, but not necessarily. If we have a round wheel rolling on a flat horizontal surface (don't look at the figure yet!), there are three possible forces
—the weight which must be vertical, pass through the center of mass, and (assuming it is a uniform wheel) pass through the point of contact; the friction, which must be parallel to the surface and pass through the point of contact; and the normal force which must be perpendicular to the surface and pass through the point of contact. If you now sum torques about the point of contact (as noted by the questioner), there are none! So, there can be no angular acceleration; if we have stipulated that the wheel does not slip, then there can be no linear acceleration either and the wheel will roll forever and no friction is required. But we all know better! A real wheel will eventually slow down. The key is that there is no such thing as a perfectly round wheel or a perfectly flat surface, one or both must be deformed. In that case, we have to think about a new kind of friction called rolling friction, the friction the wheel has because of the rolling. This is different from the static friction, and static friction may still be present still to keep the wheel from slipping. A perfectly round wheel cannot have rolling friction as I showed above, it must deform which means that there is no longer a "point (or line) of contact" but now an area of contact. Since the normal force is only constrained to act somewhere where the two are in contact, it is now possible (in fact inevitable) that this force will not act through the center of mass of the wheel. That is the whole key to answering this question. So, finally, the answer: refer to the figure where I have drawn the forces mg, N, and f. The weight is still constrained to be vertically down and pass through the center of mass (blue cross). The normal force is constrained to be vertical and act somewhere where the wheel and ground are in contact, drawn a distance d to the left. The frictional force (which now includes both static and rolling friction) is constrained to act at the surface and parallel to it. I choose a coordinate system with x to the left and y up; the axis (red cross) about which I will sum torques is at the ground directly under the center of mass and positive torque results in an angular acceleration which is positive when acceleration of the center of mass is positive (counterclockwise around the axis). All is now straightforward: ΣFx=-f=ma, ΣFy=N-mg=0, Στx=-Nd=Iα=Ia/L where I is the moment of inertia about x and L is the distance from x to x. Finally, N=mg, a=-f/m, and d=fI/(Lm2g).

Finally, a couple of real-world provisos. Of course, N is really distributed over the whole area, but the dynamics can be done by assuming it effectively acts all at one point just like we assume the weight acts all at the center of mass. And, the rolling friction might not really act at the surface of contact since it arises from the deformation of the wheel and it might not be purely horizontal since it is not directly a force due to the contact with the floor. So, there are still some idealizations in my analysis, but there are always idealizations when dealing with friction. And, the problem could have been equally well done assuming the ground, not the wheel was being deformed. One could also have done the analysis by summing torques about the center of mass and using the parallel axis theorem Icm=I-mL2.

I would like to acknowledge a very useful discussion over pizza with friends and colleagues Edwards, Love, Meltzer, and Anderson.


QUESTION:
This came to mind this morning as I was heating up my work coffee after putting milk into it. Here goes: Given (1) a set amount of coffee at a set hot temperature; (2) a set amount of milk at a set cold temperature; (3) a set amount of heat energy to be applied to the final product, in the form of microwaving for a set amount of time, say 30 seconds; and (4) a mug at room temperature: Which will lead to highest temperature of the final coffee/milk mixture: A. Heating the milk first then adding the coffee to it, or B. Heating the milk+coffee mixture together? Or, will the final temperature be the same regardless?

ANSWER:
I believe the final temperature should be larger for the milk mixed with the coffee and heated together. The reason is that the microwave oven will have a certain energy density inside and a larger volume should be able to absorb more energy in a set time than a smaller volume. If you had simply said add a certain amount of heat to the milk first or add that same amount of heat to the milk+coffee instead, the final temperatures would be the same.


QUESTION:
Why is the speed of sound slower in lead (25 degrees C) than fresh water when generally sound travels faster in solids than in liquids?

ANSWER:
The speed of sound is determined by two things, the elasticity of the medium and the density
ρ of the medium. How the elasticity is measured differs for liquids and solids, for solids we use Young's modulus Y and for liquids the bulk modulus B. So, the standard predictions are vwater=√(B/ρ)=√(2.2x109/1000)=1483 m/s and vlead=√(Y/ρ)=√(16x109/11.3x103)=1190 m/s. The experimental numbers are 1493 m/s and 1158 m/s, respectively, in pretty remarkable agreement with the predicted values. So, let's get qualitative. Lead is not at all elastic. Can you imagine how a bell made out of lead would sound? How well would a lead ball bounce off a hard floor? Also lead is very dense, so that also drags down the speed down. So, we should not really be surprised that the speed of sound is lower than in many other solids.


QUESTION:
What happens to ice-7 when it is exposed to conditions similar to the pressures and temperature of the Earths inner core?

ANSWER:
I don't really know what the pressures and temperatures are in there. But, since you are interested, you can get all that info and then look at the water P-T phase diagram.


QUESTION:
I read that if two oppositely charged bodies of same size,mass etc. are stuck together and a third charge is kept far away from them no net attraction or repulsion takes place. But if the third one comes close enough attraction takes place. It is because they can "see inside" one another and rearrange their charges resulting to a very strong interaction. What is this "seeing inside" and "rearrangement of charges"? It would be better if a diagram related to answer is given.

ANSWER:
So, I am imagining two balls, one positive and one negative, touching at their surface. This is referred to as a dipole charge distribution; it has zero net charge but not zero charge distribution. Obviously, if you get far away from this dipole, the field will be much weaker than if there had been some charge there. However, your statement that there is "no net attraction or repulsion" is wrong: there if you are on the + side of the dipole and far away a negative charge sees an attractive force, but it falls off like 1/r3 unlike the force for a net charge which would fall off like 1/r2. As you get closer, the force gets stronger and whether it is attractive or repulsive for a positive test charge depends on where you are. The electric field for an electric dipole is shown in the picture.


QUESTION:
I am falling from a plane at terminal velocity. I have a ball in my hand. I throw it towards the ground. What happens? Does the ball accelerate away from me, or does it simply fall beside me? What about if it was a very heavy ball such as a cannon ball compared say to a baseball or a plastic 'WalMart' pool ball?

ANSWER:
Terminal velocity is determined by the geometry of the object, the density of the fluid (air), and the mass of the object. The terminal velocity in air for an object of mass m and cross sectional area A can be roughly approximated by vt
≈√(4mg/A) (only in SI units). You and the ball have some terminal velocity together. When you depart from the ball you both have different terminal velocities since the masses and geometries have changed. Suppose it was a cannon ball. Then your net mass gets smaller and your geometry does not change much, so you will have a lower terminal velocity and slow down; the ball has a considerably smaller cross sectional area than it did when it was "part of" you and therefore a larger terminal velocity, so it will accelerate down until it reaches its new terminal velocity. Suppose it was a styrofoam ball. Then neither your net mass nor your geometry change much, so you will have about the same terminal velocity and continue falling at about the same rate; the ball has a considerably smaller mass than it did and therefore a smaller terminal velocity, so it will experience an upward acceleration and slow down until it reaches its new terminal velocity, you will overtake it and it will appear to go up (but is actually just going down more slowly).


QUESTION:
I have two questions regarding rotational kinetic energy. I know that rotational kinetic energy is defined as: KE = 1/2 * I * w^2 Where KE is the rotational kinetic energy [in units of: joules; kg*m^2/s^2], I is the Moment of inertial [in units of: kg* m^2], and w is angular velocity [in units of: radians/sec^2 or 1/s^2] w^2 denotes angular velocity squared so that the units check. If (Electron or Nuclear) Spin is defined as J, with values equal to a half integer * h/2pi, where h is Plank's constant [in units of: Joule *seconds or kg*m^2/s], #1, Is the rotational kinetic energy associated with Spin: KE = 1/2 *J *w ? (so the units check). Since V = w * R, where V is velocity, and R is the radius (or the Electron or Nucleus), so w = V/R, and since the fastest possible speed is the speed of light, c, #2, Is the greatest possible rotational kinetic energy associated with Spin: KE = 1/2 * J * c/R ? (so the units check).

ANSWER:
You are really out in left field here. You are making the mistake of trying to understand quantum phenomena using classical ideas. Also, your quantum statements are basically in incorrect. For example, if the angular momentum quantum number is J, the angular momentum is [√(J(J+1))]*h/(2
π), not J*h/(2π). Furthermore, if you try to look at electron spin classically by imagining a tiny rotating sphere you run into trouble first because you do not know the size or mass distribution, but even if you put in some reasonable guesses, you get absurd results for angular velocity or moment of inertia. Sometimes in nuclei which have rotational energy spectra it is useful to think about the moment of inertia of the nucleus, but only as a parameterization, not as a classical concept. Also noteworthy is that a rotational nucleus is found to behave more like a rotating fluid than a rotating rigid object. And, like any energy spectrum of a bound-state system, it is discrete, not continuous as in classical mechanics.


QUESTION:
If a motorcyclist is riding at 80 mph, is he facing or creating 80 mph winds behind him? My boyfriend insists he's facing 80 mph winds because his bike is moving at and 'hitting' air mass at 80 mph. I dont understand how his motion of mass would contribute to air mass in creating 'wind' in such precise and equal measurement?

ANSWER:
The concept here is called Galilean velocity addition. If you are in a car going 60 mph north and there is another car going 60 mph south, you see that other car approaching you with a speed of 120 mph. In this case, think of the air as the other car and it is at rest relative to the ground, that is it is a calm day; then if your velocity is 80 mph north, you see the air approaching you with a speed of 80 mph going south. Or, try this: if you look down at the ground, you see it going backwards, that is south, with a velocity of 80 mph, and since the air and ground are indisputably at rest with respect to each other, the air is also going 80 mph south to you.


QUESTION:
Why are only mass, length, and time changed as an object approaches the speed of light? Why not other "fundamental" properties of matter e.g., charge? Does this imply that charge is not fundamental or is somehow very different from mass, length, and time?

ANSWER:
The elegant way to put it is that some quantities are invariant under a Lorentz transformation and some are not. This means that some quantities are the same whoever measures them, some are not. Electric charge is invariant and so is the total energy of a particle. It would be my inclination to say that invariant quantities are more "fundamental" than those that are not invariant.


QUESTION:
when a hand is in contact with the wall the distance between the hand and wall tends to zero then why not the gravitational force is infinity between the hand and the palm as F=GMm/r2

ANSWER:
Let us simplify the picture: you have two hard spheres which you bring into contact with each other. The distance between their centers of mass is still the sum of their radii, not zero.


QUESTION:
I took two cups of unequal size (one can hold a higher volume than the other), filled the largest one with water almost to the top, and left the other one empty. Then I took a twisted piece of paper towel and put one end in each cup. Capillary action caused water from one cup to go to the other and the other began to fill. My question is why did the empty cup only fill up until both cups were of equal height, rather than of equal volume.

SIMILAR QUESTION:
I've noticed that when steeping a cup of tea, the fluid will follow the string up and over the teacup lip and create a puddle around the cup, if I leave it steeping too long. What is this principle known as? I vaguely recall it from Jr. High, but am too old to remember it clearly and I can't manage to word the question properly to get a "Google" response to it.

ANSWER:
Capillary action uses the meniscus force to lift the water to the highest point of your paper towel (string). Once it gets there, it can just flow downhill. So, in essence, your piece of paper (string) just becomes a siphon.


QUESTION:
my question is why is it that a projectile fired from a long-barrelled gun is subjected to less acceleration than a one fired from a short-barrelled gun? Thanks very much in advance.

ANSWER:
The average acceleration is proportional to the average force. When the gun is fired, there is an explosion behind the bullet which increases the pressure. This pressure pushes the bullet forward. But as the bullet goes down the barrel the volume gets larger and so the force gets smaller. So, the average force for a long barrel is smaller. Be sure to note that this does not necessarily mean the shorter-barrel gun will have a faster bullet; the longer barrel also has the property that the bullet is exposed to the force for a longer time.


QUESTION:
is the charge of a proton calculated to be exactly equal in magnitude with an electron? and what does it mean by saying equal magnitude?

ANSWER:
No, they are measured to be equal in magnitude to incredible accuracy, no different than 10-13%. Magnitude simply means that we discard the signs of the charges, the electron charge is -e, the magnitude of which is e.


QUESTION:
My husband when riding his vintage motorcycle had an accident when negotiating a bend (200m radius) in the wet. it was found that the road surface only had friction coefficient of 0.3. What formula and other known parameters (like weight of bike, lean angle etc) can I use to determine what speed he could have safely negotiated the bend? He was actually only doing about 40-43mph.

ANSWER:
I do not know where you got your coefficient of static friction
μs=0.3, but it can be only a rough estimate at best. I find that rubber on wet concrete can have μs over the range 0.45-0.75 and on wet asphalt over the range of 0.25-0.75. But, often there is oil which was in the road which floats up when it rains and makes it slipperier yet. I am assuming, since you didn't mention it, that the road is not banked. In that case, the weight of the bike+rider and lean angle are irrelevant. It is easy to calculate the minimum possible speed from Newton's second law, μsmg=mv2/R, so v=√(μsgR). Putting in your numbers I find v=24.3 m/s=54 mph. Evidently the coefficient of friction was less than you assumed. "Discretion is the better part of valor" (Shakespeare, Henry The Fourth, Part 1 Act 5, scene 4); in other words, be very careful on wet roads!

FOLLOWUP QUESTION:
The friction coefficient of 0.3 (equivalent to wet snow on tarmac) was obtained by pendulum testing in the wet to reflect the conditions at the time of the accident. You are correct in your assumption that the road was not banked. In fact the road comprises a slight adverse camber. Given the low friction coefficient I am having great difficulty in understanding how the calculation can show that a motorcycle (at normal lean angle) can negotiate the bend at 54mph. I would not wish to attempt to negotiate that bend on the equivalent of wet snow at 54mph even in a car!

ANSWER:
This is a very standard calculation. I figure that at 54 mph it would take 13 s to traverse a 90 degree turn for this radius circle which does not seem unreasonable to be possible. Are you pretty sure of the 200 m number? The key to this may well be the banking of the curve; if the road is banked wrong by 100, I would calculate v=15.1 m/s=34 mph, quite a significant difference.


QUESTION:
why don't electrons in an atom collide with the nucleus? Doesn't the nucleus of an atom have a positive charge? if so, then it should make sense that the electrons would be attracted to it, and not just go around it in orbitals. So yeah, I'd like to know why electrons don't fly into the nucleus, and stay in their orbitals. At the quantum level, I doubt gravity has any role in it, and even if it does, the mass of the nucleus would far succeed the mass of any individual electrons.

QUESTION:
What keeps an atom from getting smaller and smaller if an electron can lose energy and be attracted to the nucleus? What keeps electrons from moving closer and closer to the nucleus?

ANSWER:
The electrons are attracted to the nucleus, that is the force that holds them in their orbits. Your question is akin to "why don't the planets in the solar system collide with the sun?" Quantum mechanics shows that there is a lowest energy a system can have, called the ground state, and for atoms this is always above zero and if an electron "popped into" a nucleus and just sat there this would be a state with lower than the minimum allowed energy. On the other hand, the idea of orbits for atoms is handy but not really very good. The electrons should be thought of as a cloud and the density of the cloud at any location is representative of the probability of finding the electron there. This cloud extends all the way to the center and so there is a small but nonzero chance of finding the electron inside the nucleus. The picture at the right shows this cloud for the ground state of a hydrogen atom. The densest part is where the Bohr orbit would be.


QUESTION:
Just like a ball when hits a wall it bounces back, why dont people bounce back in the same way, when opposite force is exerted on them by the wall (as in case of ball)??

ANSWER:
It is because of elasticity. When two objects collide, how they move after the collision is determined by what happens to energy after the collision. If you drop a superball on the floor it rebounds almost as high as from where it was dropped; if you drop a putty ball, it does not. When the superball hits the floor, it compresses but it compresses like a spring which "stores" the energy. So the energy of the moving ball is moved into the energy of the spring. The spring now decompresses and gives the energy back to the motion of the ball and it flies back into the air. It obviously did not give all the ball's energy back since the ball did not quite rebound to where it was dropped from; the little bit of energy lost shows up as heat and sound. The putty ball also compresses when it hits but it bears no resemblance to a spring because after you compress it (which takes energy), it does not decompress and give energy back. Here, the energy of the ball's motion is used to compress the putty ball and that energy shows up as heat and sound entirely. The superball (neglecting the little amount of energy lost) is referred to as an elastic collision and the putty ball is a perfectly inelastic collision.


QUESTION:
Do you weigh more at the bottom or top of a large skyscraper? Does the mass over your head count for more than the distance? As an example, the tallest skyscraper currently is 828m tall and wieghs about 500,000 tons if it's empty, in Abu Dhabi. Assuming that you weigh more at the bottom, how dense or large would the mass above your head have to be to make you weigh less?

ANSWER:
There are two effects you need to think about:

  • Because you are a different distance from the center of the earth when you are at the top, your weight is less. This turns out to be a 0.03% difference.

  • The gravitational force due to the mass of the skyscraper is really small. I just took all the mass to be at the center to check this. Then, if you are at bottom, the gravitational force from the skyscraper mass would about 0.000003% smaller than your weight at the top which would increase your (0.03% smaller) weight at the top and decrease your weight by that amount at the bottom. So, the effects of the skyscraper are negligible compared to other effects.


QUESTION:
When we heat something, the particles in it move faster and faster till they reach the speed of light. But what would happen if we keep heating? The particles can move faster than the light, or there's a maximum value for the temperature (like a minimum: 0 Kelvin), or what?

ANSWER:
No, anything with mass can never reach the speed of light, but neither can it go any faster. I have answered this question before and the answer is that there is no upper limit. I have actually been corrected since there is an upper limit imposed by the total amount of energy in the universe, but that begs the question. If you have something, there is no limit to the amount of energy you can put into it, assuming you can get that energy somewhere. There is a minimum temperature possible, as you note, but that cannot actually ever be reached because of the laws of quantum mechanics. So the speed of particles in an object are constrained by 0<v<c corresponding to temperatures of 0<T<
∞ and the extremes are unreachable.


QUESTION:
Why electrons are used over proton for the charging of Body...??

ANSWER:
Because electrons are easily available because it takes only a little energy to remove them from atoms but protons are bound in the nuclei and require much energy to extract.


QUESTION:
Please explain why the energy of photons increase with frequency (not how by quoting E=hf) - do you have an analogy to explain the phenomenon?

ANSWER:
Maybe the easiest way to approach this is that, in quantum mechanics, momentum is associated with wavelength via p~1/
λ~f. Now, in relativity, E=pc so it follows that E~f. (Here, ~ denotes proportional to.)


QUESTION: 
how much time will it take to travel down a 1000 ft , 30 degree ilcline with a 225 lb payload on a 200 lb soap box race car and if pay load was lighter would the time be shorter with lighter load?

ANSWER:
I have dealt with this kind of question before, but maybe it is time to revisit the whole thing with one answer. First, consider the ideal situation where there is no friction of any kind. Physicists do not like English units, so I am going to convert everything to SI units: 1000 ft=305 m, 425 lb=193 kg. I assume that you do not want all the details of my calculations, just the pertinent results. The time does not depend at all on what the mass is (if friction plays no role). There is an acceleration down the incline which is a=
½g. I find that the time to the bottom is about 11.2 s and the speed at the bottom is about 54.9 m/s=123 mph. I assume you are not crazy enough to be in a soap box car going that speed, so friction must play a role. There are two kinds of friction you have to consider:

  • Friction due to the moving parts, like bearings, like the wheels rolling on the ground, like wheels moving on axels, etc. Empirically, we find that this kind of friction increases proportionally with how hard the moving parts are pressed together and this, of course, is proportional to the total weight of car plus rider. But, the force impelling the car down the hill is also proportional to the total weight, so once again, the speed at the bottom is independent of the load. There is something called the coefficient of friction which tells you how much friction force there is for a given force pressing the surfaces together. For example, if this coefficient is 0.3 for a 100 lb box moving on a horizontal floor, you would have to push with a 30 lb force to keep it moving with a constant speed. If I take 0.3 to be your coefficient of friction, your speed at the bottom would be about 26.8 m/s=60 mph. Since the acceleration is about half what it was, the time is about twice as long, 22 s.

  • For objects which are going faster than a few mph, air drag becomes important. Air friction depends on two things, the shape of the object and the speed it is going. It does not depend on the mass of the object. Actually a pretty good approximation to the magnitude of the force of air friction is F≈¼Av2 where A is the area presented to the wind (this works only if F, v, and A are in SI units). So, this force, which points up the incline just like the moving parts friction, causes the car to slow down more. But, Newton's second law says that a=F/m so that if F does not depend on m, the acceleration (amount of slowing down due to this force) gets smaller as m gets bigger. It gets a little complicated to actually calculate the time and speed for this case, but the important part is that this is the only place I can find where the weight of the whole car makes a significant difference.

So, the bottom line is that if you go fast enough for air drag to be important (and I suspect you do), the heavier of two otherwise identical cars should win.


QUESTION: 
If there is a yacht on a lake and it drops its anchor overboad. What happens to the water level in the lake. Apparently it falls. I have no idea why. Shouldn't it stay the same using common sense.

ANSWER:
This is a little complicated, so bear with me. First, you have to understand Archimedes' principle which states that if you have an object in a fluid, the fluid exerts an upward force on it and the magnitude of that force (called the buoyant force) is equal to the weight of the fluid which the object displaced. So, the boat plus anchor floating on the water have displaced a volume of water whose weight is the same as the weight of the boat plus anchor. Now, remove the anchor. Now the water is holding up only the weight of the boat, so the water will have fallen equivalent to removing water whose weight equals that of the anchor; but, the anchor is denser than water, so the volume of the water of equal weight is bigger than the volume of the anchor. Now, putting the anchor into the water, the water level will go up equivalent to adding a volume of water equal to the volume of the anchor. So the net effect has been a fall followed by a smaller rise, a net fall.


QUESTION: 
I understand that atoms are mostly empty space and that what we understand as matter "touching" is not really touching at all but the magnetic force of electrons repelling each other. My question is: if all this is true, how do we get bitten by mosquitoes or ticks or anything for that matter? How can something penetrate us if this "touching" sensation is actually magnetic repulsion?

ANSWER:
You have it wrong. It is not a magnetic force but the repulsive electrostatic force between the electrons in "touching" atoms. But, the force is not infinite and if you push hard enough, particularly if what you are pushing has an edge or a point which can "pry apart atoms" at the surface, you can penetrate.


QUESTION: 
What is the magnetic field of a permenant bar magnet made of? I have researched this, but cannot find a clear answer. Leading to the question...How does a bar magnets strength dissapate over time? How long will my fridge magnet stay attached to my fridge? Where does the energy come from and go to?

ANSWER:
Made of? A field is a representation of a force which would be experienced by, for example, another magnet. But, it is not really just a mathematical construct, as often taught in introductory courses, but the field exists in some "real" sense. Electromagnetic fields can have energy, momentum, or angular momentum, so maybe you would like to think of just the magnetic field being "made of energy". In prinicple, a magnet should never lose its strength because the strength because it is composed of many electron magnets which are indestructible. However, they may not all stay aligned and external fields can cause them to get unaligned. However, I would expect that your refrigerator magnets will not fall off in your lifetime. Finally, it takes no energy to hold your magnets to the fridge.  Energy requires that work be done which means a force acts over a distance and the magnet is not moving and so no work is done; zero energy is used holding your magnets to the fridge.


QUESTION: 
When a ball is thrown up in the air, does it always form a parabola? Is there any possible way that the ball could go to the top and not fall back down in a symmetrical parabola shape? ( Air resistance/ wind or something could affect this I suppose, but what about in a vaccum?)

ANSWER:
The parabolic path is only an approximation. You have to assume the gravitational force on the ball is always constant and straight down and all the "straight downs" are parallel. Well, the earth is not flat and the gravitational force decreases as you go to higher and higher altitudes. But, for usual thrown balls, etc., the approximation is excellent. Much more important is air resistance which can have a very profound effect, particularly for high speeds. Anyone who has played outfield in baseball can tell you that fly balls tend to fall more or less straight down even though they were hit at a relatively small angle. But, in a vacuum, as you suggest, the non-flat earth is all that would cause a change from parabolic. Here is an extreme example: if there were no air and you could throw a ball horizonatally at a speed of about 18,000 mph, the path would be a big circle, an orbit around the earth. See Newton's mountain.


QUESTION: 
If a light bulb doesn’t operate at a maximum efficiency, then is the law of conservation of energy correct?

ANSWER:
What does "maximum efficiency" mean? If it means that 100% of the energy put into the bulb comes out as light, then no light bulb is maximally efficient. But if only some smaller percentage of light that is emitted, the "lost" energy shows up somewhere, mostly in the form of heat. Energy is conserved for any truly isolated system.


QUESTION: 
WHAT IS MATTER? I understand that matter is composed of atoms, and that atoms are made of subatomic particles. This is the answer that most physics textbooks and google searches yield. It seems like physicists are afraid to answer this question. To say that matter is made of wave-particles is no answer at all. If it's a wave then a wave of what? If a particle, a particle of what? PLEASE I MUST KNOW!! I have lost more than a few hours of sleep over this.

ANSWER:
People lose sleep over the strangest things! I am a physicist and certainly not afraid to answer this question! My definition would be that matter is something which can be at rest in some inertial frame and which has inertia, resistance to acceleration when a force is applied.


QUESTION: 
How is it that my laptop is able charge as soon as I plug it in with out it shutting off? Does it work like a car where first it runs off the battery then a relay switches it to gas?

ANSWER:
The charger essentially looks like a dc power source of the same voltage as the battery. When the charger is plugged in, it is in parallel with the battery, so the charger both charges the battery and supplies power to the computer.


QUESTION
I've been told that when a gas is cooled, it takes up less volume, and therefore its density increases. I've also been told that a gas always fills its volume (i.e. a container). These two ideas, however, seem to contradict one another. On the one hand, cooling a gas causes it to contract and decrease in volume, which would increase its density. However, on the other hand, a gas will always fill its volume, which means its volume doesn't change when cooled. Therefore, if the volume doesn't change when cooled since a gas will always fill its volume, then its density would not increase. In other words, if a gas will always fill its volume, then cooling it would not increase its density since the gas always fills its volume and therefore remains at the same volume. In a closed container, if a gas expands to fill the volume of the container, then how could cooling the gas increase its density? How could a gas fill its container/volume on the one hand, but then on the other hand, when its cooled, decrease in volume and therefore increase in density? If a gas fills its container/volume, it wouldn't increase in density when cooled since it FILLS its volume. Therefore, since it fills its volume, the volume of the gas would remain constant. Therefore, since the volume of the gas would remain constant, its density would not increase when cooled. So my question is, how could a gas increase in density when cooled if that same gas always fills the volume of the container it's in?

ANSWER:
T
hat's a pretty long question which requires only a short answer. If the amount of gas remains constant, the relation between pressure, volume, and temperature is PV/T=constant. The situation you describe has a constraint that volume remains a constant, so P/T=constant which means that as the temperature goes down the pressure goes down; and, the volume stays the same. So, if you cool a gas in a rigid container, the density remains the same.


QUESTION
I would like confirmation about the velocity of objects moving in a circle. If an object moves in a full circle at a constant speed, is the average velocity zero? One of my friends stated that it wouldn't be zero, and that I was confusing velocity with displacement. But since velocity has direction, I thought that if I were to divide the circle into, say, ten pieces, then the velocities of each independant piece of the circle would cancel each other out because the two segments of the circle that are opposite each other would have opposite signs because they are going in opposite directions. Can you help us settle this?

ANSWER:
The average velocity is a vector and is the displacement vector divided by the elapsed time. Since the displacement is zero, the average velocity is zero. The average speed is not zero but is the circumference of the circle divided by the elapsed time.


QUESTION: 
Does cold radiates to your body if you open the door of a freezer? If so, how?

ANSWER:
Everything radiates heat. Nothing "radiates cold". The rate at which energy is radiated is determined by the temperature of the object. Your body radiates heat faster than the freezer does, so the freezer absorbs more of your radiation than you do its radiation and you feel colder because heat has left your body.


QUESTION: 
Skydiver free falls from plane with a loaded gun, and shoots gun while free falling. Does the bullet being fired travel downward any quicker then say a bullet travelling just being dropped? Well, during a heated discussion with co-workers, I was trying to explain how I thought the bullet fired downwards would not go any faster than a bullet just dropped by someone skydiving

ANSWER:
OK, let's say that two skydivers are side by side each falling with the same speed. Because of air drag, both bullets will eventually both reach "terminal velocity" which will be the same for both (assuming enough time passes before the ground is reached). But the shot bullet will have a big head start over the dropped bullet. The shot bullet will begin with a speed above the terminal velocity and slow down until it reaches it. The dropped bullet will begin with a speed below the terminal velocity and speed up until it reaches it. Certainly, the shot bullet will reach the ground first.


QUESTION: 
if i where riding a bike with a flashlight on the front end and i was going 10mph and knowing that the speed of light is constant wouldnt that mean that the light had to slow down 10 mph ?

ANSWER:
One of the basic foundations of the theory of special relativity is that the speed of light is constant regardless of the motions of either the source or the observers. You will find several questions on the FAQ page which address your question.


To see questions and answers from longer ago, link here.