QUESTION:
I have an odd question I can't stop thinking about;
I'd love some help, and I'm doing everything I can to
find an answer to this, including sending messages to
multiple people that may carry the ability to help me.
No this is not a botted message, I'm writing this in
person I assure you. taking c as the speed of light.
Now, considering an Object moving away from me to my
right at 60% of c, if it emits a light towards me that
light will travel towards me at 100% of c itself. Now
the light being emitted towards me has a relative
velocity of 160% of c to the object it's emitting the
light from. Now let us name the previous object in
question object D, and let us take another object;
object B and say it's moving away from me to my left at
60% of c. Now the light being emitted from object D is
traveling at c relative to me, and 60% of c relative to
object B, and the light is also traveling towards object
B, and we can safely assume it's bound to reach object B
at some point. This is from my perspective as an
observer (which we shall state as an observer on object
A) at rest relative to objects B and D. Now as Special
Relativity states, as long as an object is moving at a
constant velocity it's safe to assume in relation to
another object that it very well may be at rest and the
other object may be moving. Knowing Special Relativity
doesn't constrain relative speed if we shift our frame
of reference to object B, we observe object A moving
away from the speed of light from object B but we do not
observe object D or any light emitted from it as it is
moving away from us faster than the speed of light (I'm
going to assume it's possible because of relative
speeds), however considering out frame of reference from
object A, we clearly saw light from object D reach
object B, here lies my confusion. What is actually
happening? Do I lack a good understanding of special
relativity hence I give rise to faulty and stupid
questions? I'm eagerly waiting for a response, please
let me know.

ANSWER: This
question should really be thrown out because it is neither concise
nor well-focused as stipulated in the site ground rules. However, I can address it concisely so I will answer it. Yes, you
do "…lack a good understanding of special relativity…"; the speed of light in vacuum is the same for all observers.
You should see an
earlier answer discussing velocity addition in
special relativity. Using v'= (u+v )/[1+(uv /c ^{2} )] for your first situation,
v=c and u =0.6c , v' =1.6c /(1+0.6)=c ;
both observers see the same speed of the light.

QUESTION:
I am curious about the difference between mass, which I uderstand to be a constant and weight which I understand to decrease as a body departs further from a gravitational epicentre.
If weight is reduced with altitude would it take less energy to launch payloads into space by first flying the vehicle to the best achievable altitude before initiating thrust to achieve orbit?

ANSWER: The difference is important to understand. For starters, the two are different types of quantities—scalars and vectors. A scalar is specified
by a magnitude; mass is a scalar, normally measured in kilograms in science; time is a scalar, normally measured in seconds; temperature, volume, area, height, are all scalar quantities. Weight, on the other
hand is a vector, a quantity which needs to have both magnitude and direction. Often vectors are thought of as scalars in everyday life; we speak of velocity as if it were a scalar, but it is not and its
magnitude is called speed: if you say the velocity of a car is 40 miles/hour going north, that is a vector. If you say the car has a speed of 40 miles/hour, that is a scalar. Weight is a vector which has a
magnitude equal to the mass (scalar) times the gravitational field (vector). Note that the product of a vector and a scalar is a vector. The field due to a mass is a vector pointing toward the mass and it has
the units of Newtons (N) per kilogram (kg) so the weight's units are N (or pounds in Imperial units). Although I normally say that my weight is 210 pounds, that is incorrect; I should say 210 lb toward the
center of the earth (or 210 lb vertically down). You are right that you get more bang for the buck if the field is smaller, the altitudes to which airplanes (from which you might want to launch a rocket)
can fly do not really have a big reduction of field strength. I believe what you gain most at high altitudes is not having to plow through the atmosphere on your way to space.

QUESTION:
Could we detect or see a craft traveling close to light
speed as it past through our system. Or would there be
no way to detect it or even notice it?

ANSWER: The distance between the orbits of Jupiter and Saturn is about 402 million miles. I calculate the time for something traveling at a speed very
close to the speed of light to traverse that distance is about 36 minutes.
I see no reason why that could not be easily detected.

QUESTION:
I am attempting to figure out the pressure or impact the following would exert. Picture a square plate (filter plate), 48" X 48", 2" Thick, 125lb... These plates are, hanging, supported by two top-mounted angle iron bars. The plate supports have rollers at each end and the plates while hanging can be slid for cleaning. The Question: If one of that support (hanging the plate) would fail, and the plate would swing (one holder still intact) approximately 2 ft. and strike an object; what impact pressure and strike force would be applied? Summary: Square plate, appx. 125lbs. swinging appx. 4 ft., what would the impact be when the plate strikes an object?

ANSWER: I am not sure that I get the picture from your description. Seems like the plate rotates about a horizontal axis along one edge
and swings until it is vertical as shown edgewise in my sketch. Is this right? I will proceed assuming that this is right. What I can do, assuming that there is negligible
friction of the pivot and the plate, is calculate the angular velocity of the plate when it is vertical, how fast it is rotating. However, I should warn you that it is impossible to answer your question
because the answer depends on the nature of the collision between the plate and whatever it is colliding with. Let me give you a couple of examples to demonstrate that your question cannot
be answered without a lot more information. Pressure is the force divided by area. So if a small object is struck and experiences some force
F, it will experience a larger pressure than
a large object experiencing the same force
would . Also,
note that the speed of the plate close to the
pivot but gets larger farther away from the pivot;
therefore, the pressure during colliosion increases as you get farther from the pivot. Now, how do you compute the force? When the plate hits some object, the object experiences some average force
which is equal to the change in the linear momentum Δmv (its mass m times its speed v ) divided by the time t the collision acted, F =Δmv /t .
To illustrate, imagine two balls, each having a mass m , one hard like a billiard ball and another which is very squishy. When the plate strikes the billiard ball, the ball will take off almost immediately, in a very
short time, so the average force over that time will very large. When the plate strikes the squishy ball, they will remain in contact far longer so the average force will be much smaller. This also
is why if you fall onto concrete you get hurt much more than if you fall onto a mattress.
Given the fact that to answer your question, even approximately, is impossible without more information, I will stop now.

FOLLOWUP QUESTION:
Let me give you just a little more information if I may;
Here's the problem
I have a worker standing on a work platform, cleaning, and spraying "Filter Plates", these are plates on a large Water Treatment Filter Press. The plates are "hung" by two top-mounted hangers. They are in line with each other, and cascade, pressed to squeeze out water and contaminants; about 20 plates on the unit.
Concern: If (and it has happened) one of the plate "Hanging Brackets" breaks, get's loosen the plate may lose its hanging capability, not completely fall since the other hanger is intact, but Swing Out and possibly strike the worker while they are on the work platform.

REPLY: So, we have a uniform 4x4 square hanging down vertically from supports approximately at the two corners at the top. One corner fails and the square rotates about a horizontal axis through one corner and normal to the plate? Keep in mind that, as I explained in the original answer, it is not possible to simply associate a force with some moving object, you must have information about what it hits and how. Therefore you can only expect an order-of-magnitude approximate calculation making assumptions about the collision time and the angle the square has rotated through before hitting the worker. Your picture shows the square to have not yet rotated through 45^{°} and the corner hitting the worker approximately in his trunk. Are those approximately true? Would it matter if I did my approximate calculation assuming 45^{°} where the square would have its greatest speed?

FOLLOWUP QUESTION:
Subsequent exchanges with the questioner brought up the following important point: The rotating stops on its own by the time it reaches the bottom where, without friction, it would be
moving the fastest. This means that friction plays a very important role in this problem. So I have concluded that the best approach is more qualitative.

ANSWER: The figure shows the plate swinging from one support only.
L =4 ft and W =125 lb are the size and weight of the square. The angle θ denotes how far the plate
plate has rotated from its initial position, 45°; when θ =0° the center of the plate is directly below the suspension point. The length of a diagonal of the square is 2L /√2
and the weight acts at the center of the square (assuming uniform mass distribution).
I am interested in what horizontal force (shown as F in the figure) must be applied to hold the plate from rotating. I am ignoring friction for now. Summing torques to zero about the
pivot point I find that F=W [sin(θ )/sin(θ +45°)/√2].
The graph shows F plotted as a function of θ ; the required force is modest, 62.5
lb at the largest, and less than 40 lb at smaller angles where striking a worker is more likely. Furthermore, if it were moving
at smaller angles it would be going slowly because of
the frictional force; also the frictional torque would
be opposing the torque due to the weight and reduce the
necessary force F needed to bring the system
into equilibrium. If it is going slowly and the torque necessary to stop it is rather small, I would conclude that a worker could rather
easly just reach up and grab it to stop it. My only reservation would be that the suspension point at the pivot is not designed to be a pivot and might not always behave the same; if the friction were such that it would pass
the equilbrium position with a significant velocity, it could be more dangerous.

ADDED
THOUGHTS:
I have done some additional calculations which I believe will shed a little light on this problem.
For the convenience of the reader I have copied the figure from earlier in this Q&A (ignore F for the following discussion) The angle θ (t ) is the angle between the vertical and the line from the pivot to the center of mass of the
square filter. The torque due to the weight is τ _{W} =-[WL sinθ ]/√2.
(The negative sign is because the torque due to the weight will cause a clockwise acceleration oppsite the direction of positive θ which is counterclockwise.) I will now add a constant frictional torque due to friction in the attachment
which is being pivoted about, τ_{f} . When the other attachment fails, it can be seen that θ (t= 0)=45°=π /4.
The initial value of the angular velocity is θ' (t= 0)=0. Now, Newton's second law for rotational motion is that the moment of inertia
I times the angular acceleration equals the sum of the torques
about the rotation pivot, I θ" =τ _{W} +τ_{f} ; The moment of inertia of a square about an axis perpendicular to the square
at one corner is I =2ML ^{2} /3. Now, this equation is transendental,
i.e . it cannot be solved analytically. So I used a computer program to solve it numerically.

First it is instructive to solve the differential equation without friction. The figure below shows the the angle (called
y here) as a function of t . Note that the filter swings like a pendulum until it reaches -π /4 in about 3/4
of a second, all the way to the other side. Now, although it is momentarily at rest, there is still
a torque on it due to the weight which will start accelerating it in the counterclockwise direction back to
+π /4 and so on. The numbers 56.3 kg·m^{2
} and 1058 N·m are the moment of inertia and maximum torque (at θ =π /2) in SI units,
respectively.

Finally I will add a constant frictional torque. The final result is shown in the figure below. The frictional torque,
394 N·m, was determined by adjusting it until the angle was zero when the angular speed first reached
zero. Now, the center of mass is directly below the
pivot so the weight no longer exerts a torque and the filter is at rest; since the friction will only exert a force if the filter is moving, the filter now stops dead.

QUESTION:
Holding a car FOB against your head will increase its effective range. For example, if you are trying to lock/ unlock your car from a distance and the FOB battery isn't strong enough to make a strong enough signal to reach your car, you can hold your car key against your head... which extends the effective distance of the FOB by several/ dozens of yards, and may allow you to then lock unlock your car if you are just out of your FOB's range.
There are many websites which document this. I know it sounds crazy, but it's real. What is your opinion regarding how this trick amplifies the FOB signal?

ANSWER: It is not my "opinion", it definitely works and the reason is very well understood. The explanation is fairly long and technical
and involves the wavelength of the beam from the FOB, the size of your body, the size of your head, resonance, and the dipolar property of water molecules. There is a pretty clear yet entertaining video
which you can watch at this
link.

QUESTION:
Hello, this is to settle an argument between my brother and I and we can't find an answer, so I hope you can help us.
There are two soda cans cold out of the fridge, both at the same temperature and sitting in the same environment (say on a coffee table). One can has been opened and the other remains sealed. Does the can that has not been opened warm faster because its still sealed and under pressure.

ANSWER: Let me first address how heat is transmitted to and through the interior of the unopened can. We generally think of heat transfer as occuring via three ways: radiation, conduction, and
convection.

The can has two surfaces, the bottom and side, where heat may be conducted through the aluminum directly to the soda; the third surface, the top, has aluminum in contact with the gas at the top
through which which heat may be conducted to the surface of the soda. The rate (R ) of heat transfer through a medium of thickness t with a temperature
difference ΔT is proportional to
ΔT

and inversely proportional to t ,
R=σ ΔT /t , where σ is the conductivity; clearly, since the conductivity of
the gas is much smaller than for the aluminum and its
thickness much larger, much less heat is conducted through
the top.
Now, the soda near the inner surface is warmer than
it is inside, so now the heat could conduct into the cooler volume. But, in a fluid, heat is much more efficiently distributed via convection, currents in the
fluid. Similarly, heat through the gas at the top of the can will occur much more efficiently by convection.
I don't believe that radiation plays a particularly important role in the heating of the soda.
So, now, what changes when the top is opened? As you noted, the pressure in the gas will decrease meaning a smaller density and therefore a smaller conductivity. But, convection will play the major role
in the gas and I believe that the gas will quickly equilibrate with the outside air; then, the temperature at the top surface of the soda will be the same as outside air. Also, the soda in the open can will
be losing its carbonation, but since there is only about 0.04 oz of CO_{2} in a 12 oz soda, I would expect no significant changes in the properties of the soda as it loses the CO_{2} .

So, I expect the open can to warm up faster. But, I am not absolutely positive and suspect that there will not be a big difference.
Here is what I would do: get an instant read thermometer and put it in the open can; then wait until the soda gets to about ¾ of the room temperature and quickly open the closed can and measure the temperature. If you
do this experiment, please let me know the result!

QUESTION:
My question is: assuming space was a perfect vacuum, (for simplification purposes) as light travels through it, and then slows down (however minutely)by entering a planet's atmosphere, once it passes through the amosphere and re-enters a vacuum, does it resume (accelerate to) its original speed or does it remain at the slower speed caused by the atmosphere?

ANSWER:
In any medium light has a speed v , a frequency f , and a wavelength λ . The energy of any single photon is E=hf where h is Planck's constant; also, v , λ , and f are related by v=λf .
The energy of any photon cannot change, so
f remains constant regardless of v . Therefore when the light moves from the atmosphere to a vacuum, its wavelength increases so that v=c where
c is the speed of light in a vacuum.

QUESTION:
Calculate the kinetic energy of a 5,000 kg meteoroid traveling at 11.2 km s-1. If it were to impact the surface of Earth at this velocity, what would be the equivalent strength of the explosion using the units of Tons of TNT?

ANSWER:
The speed is very small compared to the speed of light (c =3x10^{8} m/s), so classical mechanics can be used. The kinetic energy is K =½mv ^{2} =3.14x10^{11} J.
1 Joule (J) is 2.39x10^{-10 } tons of tnt, so the
energy released is 3.14x10^{11} x2.39x10^{-10} =75
tons of tnt.

QUESTION:
I watched the "trip to infinity" doc on netflix and there was a part that said if you put an apple in a box, that apple will eventually decompose but the trapped energy will eventually create something else. Can the same concept be applied to humans and rebirth? Is our body just energy that takes another shape eventually after we die?

ANSWER:
Assuming you have a box where nothing can go in or out, you would have an isolated system consisting of the apple and whatever air is inside. If there are no bacteria present there
would be no mechanism for decomposition. Even so, the water in the apple would evaporate out, eventually drying the apple. One apparent energy transfer would be in the form of heat and it would warm up inside.
But, the heat can't escape. When the water evaporates, some chemistry might happen inside the apple releasing heat energy and changing the composition of what is left of the apple. Etc ., etc ., etc .
But you can be sure that the apple will not eventually become a mouse or a peach or even a new apple. Putting you in the box would eventually result in the same fate as the apple, but you can be sure
that rebirth is not in the cards. Of course, there is no such thing as a perfect isolating box and heat, the main result of the apple's demise, would eventually leak out and radiate away until the inside and outside
of the box were in thermal equilibrium.

QUESTION:
In general science class years ago I learned that
when a balloon is inflated and the stem released, the
pressure of the air in the balloon pressing against the
front side opposite the stem is greater than the ambient
air pressure in front of the balloon and since the
pressure at the stem is being released, the balloon
moves toward the lower pressure area ahead of it. This
is not hard to understand for a balloon or rocket.
However, I don't understand how this principle works in
a jet engine. Since the forward end of a jet
(particularly a ram jet) is open also, what is the
interior pressure pushing against to move the jet
forward?

ANSWER:
Frankly, I do not particularly like this way to explain rocket propulsion. It is much more easily explained using Newton's third law, if object A exerts a force on object B, object B exerts an equal and opposite force
on object B. Think of each molecule of the ejected gas: for it to get propelled out, the engine must exert a backward force on it; therefore this molecule of the gas exerts an equal force on the engine
but in the forward direction. I suggest that you read two earier answers, (1 ) and (2 ): The first discusses rockets and such, the second has a discussion on how a jet engine works.

QUESTION:
I have read somewhere that only the Earth below, not the Earth above, exerts a gravitational force on a deeply buried piece of matter. What does this statement mean? We know all masses exert gravitational force on each other. Then why upper part of earth does not exert gravitational force on deeply buried objects below?

ANSWER:
First, for this statement to be true, the earth must be a sphere and spherically symmetric. (Spherically symmetric means that, at a particular distanc from the center of the earth,
the density is the same regardless of the direction of that radius vector. So, for example, if you are 100 miles toward the north pole, the south pole, or the equator, the density is the same at all
three of those positions.) Consider first a tiny piece of mass right at the center of the earth. you should intuitively see that it will experience no force because for every little piece of mass inside
the earth exerting a force on the centered mass there is mass on the other side of the earth which exerts a force of the same magnitude pointing in the opposite direction, so the force due to these
two "earth above" pieces of the earth cancel to zero. Similarly every other piece of the earth has
a partner exerting an equal and opposite force on the centered mass. (This has nothing to do with Newton's third law!)
The net force on a small piece of the earth's mass anywhere but the center is considerably harder to calculate and beyond the scope of this site. The result however is that the net force on any small piece of
earth is equal to the force which a point mass equal to the mass of the "earth below" located at the center of the earth.
To learn more about details, learn about Gauss's Law.

QUESTION:
From the perspective of light (hypothetically), how much time passes approximately for it to get from andromeda to us, 2.54 million light years away.
I know the obvious answer (2.54 million years) but I am wondering whether time dilation due to velocity occurs, and whether the distance from any gravity(Mass) has any additional time dilating effects.

ANSWER:
The length of time that anything happens depends on the observer. Yes, the earth will see the light
from Andromeda arriving after a trip of 2.54 million years. But, something moving relative to the
earth will see it take some different time. For example, suppose there is a space ship moving from the earth toward Andromeda with a speed 80% the speed of light. An observer on the ship will see the distance
from earth to Andromeda to be shrunken, 2.54x√(1-0.8^{2} )=1.52 million light years and therefore the light takes1.52x10^{6} years to make the trip in his frame. If you are asking (as many have) how long it
takes in the photon's frame of reference, a photon has n perspective because it cannot have any kind of clock it carries. See earlier
an answer
and an earlier yet answer
linked to.

QUESTION:
If you have a house and it's 32F outside, do you really save on heating fuel if you set your thermostat on 68 instead of 72? I hear them suggesting that all the time but what I recall from HS physics class is that if the furnace turns on at 4 degrees below the setting it's going to take x-btu to raise it that 4 degree range regardless.

ANSWER:
You recall wrong. Fewer BTU are required to increase temperature 4° starting at 50° than starting from 60° because heat leaks out of the house faster at higher
temperatures. So let me try to convince you that you consume less fuel if you keep your thermostat at a low temperature. Here is my scenario: when I awaken in the morning when the temperature is
32° outside the temperature is 60° inside because that is what I set my thermostat to when I go to sleep. Now I turn my thermostat to 68°, it will take a certain amount of energy to raise the inside temperature
to the set temperature; however, if I set it to 72°, it will take more energy to raise it to 72° because it has to raise it to 68° on the way. In order to hold the temperature to any constant
value the furnace must replace the energy which is leaking from the house, through walls, ceilings, floors, doors, windows, etc . It turns out that the rate that
heat conducts through a barrier is
proportional to the temperature difference across the barrier. Therefore, the heat leaks out of the house faster if the inside temperature is 72° than it does when it is 68°; therefore,
more heat from your furnace is required for the 72° thermostat setting.

QUESTION:
Am I right in assuming the diagram they normally use to illustrate how
mass bends spacetime, usually depicted as a sheet with bowling
ball in the middle of it, which creates a cone like depression in the sheet representing the curvature of spacetime created by mass that draws in other objects is completely oversimplified and inaccurate because no matter what direction an object is approaching a planet from, it is drawn into the gravity, so the curvature must be a sphere around the planet rather than a cone like depression underneath it ? I have never taken physics so please excuse ignorance if I am completely wrong.

ANSWER:
See the
FAQ page. Search for "trampoline model".

QUESTION:
So if I bowl a 14lb ball at a speed of 24mph, how much would the force weight be when the ball impacts on the pins. I was talking with a friend about this while bowling and I was telling him that the force and the impact would be much greater then bowling it at like 1mph and hitting you. I just don't know how to calculate it.

ANSWER:
It is very cumberson to do physics using imperial units, so I will switch to SI (metric) units and switch back later. A 14 lb ball has a mass of about 6.4 kg; 24 mph is about 10.7 m/s; 1 mph
is about 4.5 m/s; we will also want the mass of a bowling pin, about 1.5 kg and the mass of 200 lb man, about 91 kg. I will first consider the bowling ball colliding
elastically and head-on (both traveling in the same direction as the incoming ball) with the pin first. With no extra detail here, I find that after the collision the bowling ball has a
speed of about 6.64 m/s (14.9 mph) and the pin has a speed of about 17.3 m/s (39 mph). How did this happen? During some time t the two were in contact exerting equal and opposite forces on each other, the
ball slowing down and the pin speeding up. The average force experienced by each has a magnitude I will call F. Over the time t the change in momentum (mass times velocity) of the pin (which is
(17.3x1.5-0)=26 kg·m/s
is equal to Ft which is called the impulse. So
the force of the ball on the pin is F =26/t. I don't know what the
time of the collision is but I know it is small; let's just say that it's on the order of a hundreth of a second,
t =0.01 s. Then =0.26/0.01=2600 N=585 lb. Now consider the ball hitting the gut of the 200 lb=91 kg man. This is not an elastic collision because I would assume that the ball and man will remain
stuck together after both are going with the same velocity which would be about [4.5x6.4/(91+6.4)]v =0.3 m/s. Again, I do not know how long the collision took but it is a lot longer than a hundreth of a
second, let's estimate that t =0.5 s. Now the impulse is 0.5F =0.3x91=27.3 kg·m/s, so F =27.3/0.5=54.6 N=12.3 lb. The bottom line is that the force of the ball on the pin, 585 lb,
is much larger than the force, 54.6 lb, of the slower ball on the man, mainly because the duration of the collision is so much shorter for the ball-pin collision.

QUESTION:
So i'm having a bit of a heated scientific discussion with a pal the minute. My question is if you threw (not dropped) a light object and a heavy object with the same force in the direction of the gravitational pull, which object would express more acceleration if there is any difference?

ANSWER:
During the time you are exerting the force
F (presumably straight down) on each object, the total force on it would be mg+F where m is the mass of the object and g is the acceleration due
to gravity; mg is the weight. Newton's second law says that the net force is equal to mass times acceleration, so the acceleration a is a=g+F /m . So, while the force acts , the
heavier mass has the smaller acceleration. Once the force stops when it leaves your hand the accelerations become identical,
g , regardless of the fact that they may have different speeds when released. All this
ignores air drag.

QUESTION:
as a mechanically-biased person but untrained physicist, who has an interest in all things scientific, one thing has been puzzling me lately.
When particle physicists talk of manipulating particles (say, in the LHC, or discussing entanglement, or even in the slit experiment) - how do they isloate just one or two particles? And what might the particles be? Even a single photon must involve alot of physical process just to catch it and use it. How do they do the 'nitty gritty' 'hands-on' stuff?
I guess it's a mechanical and not a theoretical concern that I'm having trouble with!

ANSWER:
I will talk first about photons. Every time one excited atom decays to a lower-energy state, one photon comes out. The same is true of an excited nucleus. Exciting atoms and looking at
their spectra is 19th century physics. It is often done by passing an electric current throught a gas; the electrons excite the atoms of the gas which emits photons and detectors see them, one by one.
But wait a minute, where do the electrons come from. It turns out that electrons are very easy to get: simply take a tungsten filiment and heat it up until it glows red or white. Electrons stream off
the filament and electric and magnetic fields can manipulate them. Heavier things need what are called sources to strip electrons off of atoms and the ionized atoms again can be manipulated with EM fields.
Often those ions (simply protons if the ionized atoms are hydrogen) are accelerated and "shot" at a target, often a thin foil or container of a gas. Then they pass through the tartet and some interact with
the target atoms and they may bounce out having been scattered by the target atoms. The scattered particles are then detected, one by one.

QUESTION:
The wave solution to the Photoelectric effect.
Hi there.
I was recently examining the relationship between the work function of a material and its threshold wavelength. It was clear to me that the relationship is expressed as:
(λW)^{2} = c/2
Where λ is the threshold wavelength, W is the work function in Evs, and c is the speed of light. However, I am unable to find the name of this solution online.
Do know what this relationship is called, and a link that explains it?

ANSWER:
The equation you are asking me the name of cannot be any meaningful equation at all because it is not dimensionally correct. The work function has dimensions of energy
which would be mass·(length/time)^{2} =M·L^{2} /T^{2} , so the left side of your equation has dimensions (L·(M·L^{2} /T^{2} ))^{2} =M^{2} ·L^{6} /T^{4}
and the right side is just a speed L/T. The fundamental equation you seek for the photoelectric effect is that the work function, the minimum energy required to remove one electron from the surface,
is equal to the energy of a photon of frequency f
which is Plank's constant h times f , W=hf . If you would rather use the wavelength
λ than f , note that λf=c , so W=hc/λ .
I am not aware of any name this equation has. Once you have defined what
W is, the only physics is Einstein's postulate that the energy of a photon is
hf .

QUESTION:
According to a video game, a 20kg iron bullet traveling at 1.3% the speed of light will impact a surface with the force of a 38 kiloton bomb when sent through the vacuum of space. But is it true?

ANSWER:
First, it is not force but energy delivered that you want to talk about. What is the energy delivered by this "bullet"? We can use Newtonian physics to detrmine the answer because
this speed is very small compared to the speed of light. So the speed of the bullet is
v =0.013x3x10^{8} =3.9x10^{6} m/s. So the energy E can be written as E =½mv ^{2} =1.52x10^{14} J
where m =20 kg is the mass of the bullet.
Now, 1 Joule (J) is 2.39x10^{-13} kilotons of tnt, so
E =1.52x10^{14} x2.39x10^{-13} =36.4 kiloton.
This is pretty close to 38 and video game guys may have
used a slightly larger value of the conversion factor.

QUESTION:
I am wanting to build a greenhouse/garden with a big water tank buried underground. I would like to add a spigot at slightly above ground level where I can attach a hose or fill a bucket. Would the pressure from gravity on the top of the water in the tank be enough to push water out of a pipe say half a meter above the top of the tank? Let’s say a 40 mm pipe.

ANSWER:
Absolutely not! What you refer to as "pressure from gravity" is actually pressure from the atmosphere and atmospheric pressure is virtually the same at the hose as it is at the
tank. If you want to have the tank buried you will need a pump
pull up the water. If all you generally need is a bucketfull, just get a hand pump like farm houses all used to use to bring up water from a well.

QUESTION:
Would less energy be required to launch a spacecraft by having in take off like a plane, using a runway/ramp?
Using retractable/eject-able wings for lift seems like you would conserve energy, as opposed to launching straight up (fighting gravity without benefit of atmospheric lift provided by wings).
A pilot buddy of mine said the energy used would be the same, regardless of method used.

ANSWER:
I expect that lift from wings would be trivially small compared with the force needed to accelerate up to orbital speed. Also, because of energy lost to air drag, it
is required that the rocket get out of the asmosphere as quickly as possible. Also, because most rockets big enough to put significant payloads into space would be much heavier than any airplane, there would be
an enormous engineering problem to get wings big enough and strong enough to provide lift just to keep it from falling, let alone rising.
And, I think your buddy is wrong. Space rockets get up
to high speeds very rapidly and the energy loss due to air drag would be significant.

QUESTION:
Does an electric car with a direct drive DC motor use any energy to stay stationary on a hill? The car has a "hold mode" which prevents the car from moving downhill when stopped.

ANSWER:
I am pretty sure if the ignition (do they call it that for electric vehicles?) is on that some energy is being used by various electric devices (radio, fan, AC, lights, etc .) but that probably all
comes from the battery as opposed to the motor itself.

QUESTION:
Comparing gas combustion vehicles to electric vehicles, if a Camry/Accord weighs 3300 lbs and maybe add 100 lbs for gasoline and if an electric battery weighs 1000 lbs making an electric vehicle weigh about 4300 lbs, what is joules per mile to move a gasoline combustion car vs joules per mile to move and electric vehicle?
Electric vehicles weighing so much more I just can't see how they are more energy efficient. Am I missing something. Also wouldn't you agree that electric vehicles are by and large powered by hydrocarbons?

ANSWER:
Yes, you are missing something. Electric and internal combustion engines have very different efficiencies. Electric motors can have efficiencies up to about 85% whereas internal combustion
engines are more like 40% efficient. This means that an internal combustion engine wastes approximately 4 times as much energy to heat than its electric counterpart (60%/15%). A reasonable estimate of how much
of the energy is lost because of the weight to keep a car running at a constant speed would be that the amount would be proportional to the weight; so the case you give would be that the internal combustion car
would waste about 0.77 (3300/4300) that of the electric car due to frictional forces for similarly designed cars.

So there is no question that the electric car would be the hands-down winner in a Joule/mile comparison. However, this is only
part of the story. To make a complete inventory is a
much more complicated issue. As you note, most electricity is still generated using hydrocarbon sources—oil, natural gas, and coal; as time goes on, this will change to more renewable sources. Also, the energy used
to manufacture the batteries and transport them to the vehicle is not negligible. Also, transporting gasoline to the consumers is responsible for carbon emissions whereas transporting the electric power is not.

QUESTION:
Is there any way we can measure surface tension of liquids at a school laboratory or our home (in other words, without any special equipment)?

ANSWER:
First I want to apologize to the questioner for taking so long to answer this question; when I first researched finding a way I was unsuccessful and put it aside for the time
being, just returning it to today. Today I found an
excellent article giving three ways using only easily available equipment.

QUESTION:
I'm writing a book where at one point my main character, in a single pilot private airplane, falls from the sky before she catches herself. I need to know how many G forces would affect her as she falls. I have a few numbers but I'm not sure what to do with them!
Her plane weighs about 7500 lbs. She falls for 45 seconds. The average commercial plane weighs about 530000 lbs and receives between 6-12 G forces as it falls. How many G forces does my small aircraft receive?

ANSWER:
I find your question very puzzling. Let's think about the falling airplanes. If you drop an airplane it will begin to fall with the acceleration due to gravity g which is about g =32 ft/s^{2} which
means that during each second it falls, it speeds up by an amount 32 ft/s. If that airplane's weight is the only force on it, it experiences zero G force; it's like someone in an elevator where the
cable has snapped and that person feels "weightless". But the weight is not the only force on the plane, there is also an air-drag force upwards; so, when the plane has acquired some downward speed it has
an acceleration less than g so the net force on the plane is a little bigger than zero G force. Since the drag force gets bigger as the speed down gets bigger, eventually the plane will be
falling with some constant speed called the terminal velocity . Now someone on the plane will feel a G force of exactly 1.0 G. So I have no idea how that 6-12 G for a falling airplane could be right unless
the plane was in a rapid spin as well as falling so that the forces on the wings could very well exceed their weight resulting on forces bigger than 1.0 G. Also the magnitude of the air drag depends on the orientation
of the plane—a plane falling nose down would have a much smaller air drag at some particular speed than one falling belly down. So, I don't know what you are hoping to achieve in your book, but getting a huge
force on a passenger in a plane dropping out of the sky is not going to happen. Since you indicate that she manages to get the plane to stop falling at some point, she might experience a large G force during
that time since that would require her achieving a very large acceleration upwards during the maneuver to stop her fall.

QUESTION:
Is Galileo law of free fall true?

ANSWER:
It is true only if air drag is neglected. If there were no air it would be exactly true. Imagine dropping a golf ball and a sheet of paper; they obviously do not experience
the same accelerations as they fall to the ground.

QUESTION:
Will Rotational inertia increase as a gymnast moves from an extended position to a tucked position?

ANSWER:
No, rotational inertia (usually called moment of inertia, I , decreases as mass of the object is moved closer to the axis of rotation. (See the question immediately following yours.)
Since the angular momentum (moment of inertia times angular speed,
ω ) is approximately conserved in this
situation—Iω =constant—reducing I will cause ω to increase.

QUESTION:
We see that ice skaters spin faster then they draw in their arms. What happens if then simultaneously draw one arm in while extending the other? Would this negate the effect?
Obviously the center or gravity would shift, but assuming this was compensated for would there be any change in rotational speed?

ANSWER:
The wording of your question is confusing. She starts spinning with her arms out. One thing she can do is pull both in which causes her to spin faster. If she only pulls one arm in
she will also spin faster but not as fast as when she pulls in both. Ignoring friction, if she puts both arms back out she will slow down to the spin speed at the beginning. The physics reason is that
if an object is spinning about its axis you can speed it up by making its moment of inertia smaller; if the object is spinning about its axis you can slow it down by making its moment of inertia larger. The
moment of inertia is a quantity which measures how close the mass of the object is to the axis of rotation; if the skater could pull in not just her arms but suck her whole body in so that it is all say 5 cm in diameter
it would spin enormously fast! I have seen skaters begin the spin with one leg extended out which gives even more spin as she moves the mass of arms and that leg in closer to the rotation axis.

QUESTION:
I have gotten into knife sharpening lately, which has Got me thinking about steels’ structure. Can steel, in theory, get as Sharp as one molecule at the edge?
Can some steels, in theory, get sharper than others, Because of different structual grits?

ANSWER:
I am not going to get into the steel issues because knife and sword making and its sharpening is almost an art form. Also, you have to ask what the knife is for; generally judged the sharpest
is one with a 10° bevel on the edge. But you are probably interested in the limits. I have read that a diamond can be sharpened to the degree where the edge is about 1 nm (10^{-9} m) wide; the diameter of an atom
is about 0.2 nm, so the edge is around 5 atoms wide. Obsidian, favored for arrowheads, can be about 3 nm wide, about 15 atoms.

QUESTION:
My understanding is if I drop an object, the
object is stationary and it the earth that rises to meet
it @ 9.8 meters sec squared. It is very difficult for me
to visualize why the earth can appear to remain
stationary while it continues to expand.

ANSWER:
I don't know where you got your "understanding", but it is totally wrong. The object accelerates toward the center of the earth (vertically) at g =9.8 m/s^{2} and the earth remains
at rest. If you want to get really picky, you could say that the object accelerates downward with an acceleration
^{g} =9.8 m/s^{2} and the earth accelerates upward with an acceleration a tiny bit
larger than zero. Here is how it goes: Newton's third law states that if the earth exerts a force on the object of magnitude W (the weight which has the magnitude mg where m is the mass of the
object), then the object exerts an equal and opposite force on the earth. Newton's second law says that the acceleration of an object is equal to the force it experiences divided by its mass. So, call the mass
of the earth M and its acceleration A ; then the acceleration of the earth is A=mg/M and the acceleration of the object is a=g . The mass of the earth is
M =6x10^{24} kg so, for an object
with mass 1 kg the earth's acceleration is A =1x9.8/6x10^{24} =1.63x10^{-24} m/s^{2} ,
essentially zero!

QUESTION:
I'd like to know if solid objects move
through the different gases (hydrogen, methane, oxygen
etc) differently. I'm 51 yes old and this isn't my
homework but I asked this same question multiple
different ways and the internet so far has yielded no
answers of any real relevance.

ANSWER:
When an object moves through a gas it experiences a drag force
F in addition to any other forces which might be acting (e.g. , gravity). The force, which acts opposite the object's velocity
v , is given approximately by F =½ρ v ^{2} CA where ρ is the density of the gas,
v is the speed, C is a constant which depends on the
shape, and A is the effective cross-sectional area the object presents to the onrushing gas.
For example, for a sphere of radius R , C =½, A =π R ^{2} , so
F =¼π R ^{2} ρv ^{2} . Clearly, the density of the different gasses will be different for a given pressure and temperature of the gas. So the answer
to your question is "yes" .

QUESTION:
I have been atempting to find a formula that calculates the mass of a planet that dose not rely on acceleration due to gravity OR a formula that dose not rely on planitary mass to calculate acceleration due to gravity, and in both situations densty is not known. i have noticed that if i find a formula that completes this calculation then the answer can then be used to determine the others value. I have found the formula
M = 4π^{2} r^{3} /gt^{2}
But to my understanding the value of (g) is acceleration due to gravity so cannot be the formula i am looking for, in an ideal answer the formula is one that could be applied to a foregn planet and insted relys on values visualy obtained; orbital period, orbital radius, planitary diamiter, planitary volume in any number or order of visualy obtained values. If the above formula is not the one i am looking for can you include the correct one please?

ANSWER:
You have run upon the problem which Newton had to struggle with—it is not possible to find the mass of a spherical gravitational source without knowing the magnitude
of the force between two point (or spherically
symmetric) masses. Newton's guess for the force between two point (or spherically symmetric) masses, M and m , separated by a distance R , was
that the magnitude of the force F is proportional to Mm /R ^{2} , or F=G Mm /R ^{2}
where G is the proportionality constant. Just knowing this results in completely deriving Kepler's laws of
planetary motion, but you could not find any masses without knowing the proportionality constant G .
Newton made a lucky guess that if the average density of the earth was between 5 and 6 times the density of water and g =9.8 m/s^{2} , then G =(6.7±0.6)x10^{-11} m^{3} ·kg^{-1} ·s^{-2} ;
this is extremely close to the modern value G =(6.67430±0.00015)x10^{-11} m^{3} ·kg^{-1} ·s^{-2} .
(See more detail
here .) This should help you with your dilemma. At
the surface of a planet of mass M and radius R the weight of a particle of m is W=mg . But the weight of a particle is the force which the planet exerts on it, i.e. , mg=GMm /R ^{2
} or
M=gR ^{2} /G .

Laws of physics generally take the form of proportionalities. To see if they are valid laws one must make the proportionality become an equation by introducing a proportionality
constant. To determine if this is a correct law, one must determine the constant by doing an experiment.

QUESTION:
In one of the Superman movies, Superman saves a plane from crashing by holding it by the nose and lowering it to the ground, in the show The Boys, a character named Homelander refuses to save a plane full of people because he claims it's impossible to save the plane without destroying it. Do you think it would be possible for a person with superpowers to hold a plane by its landing gear to slow the plane down and lower it to the ground? Yeah I'm arguing with somebody on facebook about this, I think it would be possible, they're calling me names and saying I don't understand physics, so I googled "ask a physicist" and here I am. Hope you can answer my question and I hope you enjoy thinking about it.

ANSWER:
To see a clip of what is being described here, link
here .

The site ground rules specify that questions based on unphysical assumptions (superpowers?) are not answered. In this case I feel that some discussion is warranted. One of my biggest
problems with Superman is, what does he do for
propulsion? In order to stop or start or turn there needs to be a force on him. He can't exert a force on himself. And if he pushes hard enough to stop the plane in midair he would have to exert a huge force but Newton's third
law would stipulate that the plane would exert an equal and opposite force on him; because his mass is tiny compared to the mass of the airplane, the plane would have a tiny change in its speed whereas Superman would
be pushed off with an enormous speed opposite the direction he is pushing. Also, the "skin" of an airplane is relatively thin and pushing with a huge force over a small area would punch right through it. No matter how strong
he is, there is no way this is in any way possible without violating the laws of physics. Of course, if you say it is possible because superpowers permit laws of physics to be violated, then it is possible; but if that
were the case, the discussion would be inappropriate for a physics site.

The stuff I have been talking about is discussed in more detail than my answer on this
link .

QUESTION:
I hope you can help with a slightly random one. I happened an article about the recent Nobel Prize-winning work done by Alain Aspect, John Clauser and Anton Zeilinger concerning entangled quantum states. Full disclosure, I am not an academic or even remotely versed in your world. But, as a layperson (I'm creative in advertising) it certainly caught my imagination and got me thinking.
My question relates to the experiment's use of lab-grown diamonds over natural ones. I understand their research has untold implications and possibilities, but for myself (and my wandering creative mind), it got me thinking more deeply around the
"what ifs" involving the emotional value people could put on having a pair of entangled diamonds.
For decades, natural diamonds have been the symbol of true love, but what if lab-grown diamonds could tell a greater love story, based on the idea that only they can talk to each other and feel each other's vibrations?
Knowing this, my burning question centres around - Whether entanglement is exclusive to lab-grown diamonds vs mined ones. And do the pair of diamonds need to be made at the same time? What makes them suitable for entanglement? This is what we are hoping to find out in the simplest of terms.
I'm looking into this as part of a pro-active passion project. It's a story I feel worthy of telling and I'd value any assistance or guidance. When considering the true value of lab-grown diamonds, it's not about the price or rarity, but more about the emotional worth to people. And if entanglement can only happen between the lab-grown variety over mined ones, then it's a story that could radically change people's opinions and actions when buying them.

ANSWER:
As I have admitted
before , I am not an authority on
entanglement.
However, I believe that I can answer your question quickly. Every crystal has some flaws, either presence of impurity atoms or dislocations of the crystral
structure itself; no diamond is 100% pure carbon atoms. The presence of certain impurities
in a crystal often determine its color. It turns out that a particular impurity, nitrogen, in a diamond is a particularly useful candidate for studying entanglement; called nitrogen vacancies (NV), two NVs
can be entangled and are quite stable. This is what is being entangled in the experiments you cite, not two diamonds. My guess is that lab-grown diamonds are used in these experiments because
the impurities can be controlled during the growth process.
Your idea was lovely and romantic, and I am sorry to have to shoot it down!

QUESTION:
This is an odd one recently I saw a graphic saying that the Ocean on Titan could be 100KM deep and have a 20KM ice sheet. What my question is, is what would the pressure do to water a 100KM deep? I know from my crude calucations that 1km is 100bar on earth so that means the pressure could be 10000 bar or about 90,000PSI. Would some sort of fusion occur or would any heat be generated at that depth at all?

ANSWER:
It really isn't clear what the oceans are composed of—water, methane, ammonia, mixtures, etc .—but you have made some other miscalculations so let's just do the necessary calculations
for water as an example to understand how to do it. First of all, you have apparently used the acceleration due to gravity to be about 10 m/s^{2} , just like on earth. But, the actual value of g on the surface of Titan
is only about 1.35 m/s^{2} , much lower. Since the depth you want to calculate, 120 km, is very small compared to the radius of
Titan, about 2600 km, I will assume that g =1.35 m/s^{2} is constant. (It will actually get smaller as you
go deeper.) Water is nearly incompressible so I will approximate its density ρ =10^{3
} kg/m^{3} to remain constant. Also, because the density of ice is approximately the same as water, I will just calculate the pressure at a depth
d from the surface of the ice as if it were water also.
Finally we can write P=P _{0} +ρgh where
P _{0} is atmospheric pressure at the surface, about 1.47x10^{5} N/m^{2} .
So, P =1.62x10^{8} N/m^{2} =1.62x10^{8} Pa=1620
bar=23,500 PSI. The phase diagram shown shows that for the water to not freeze at high pressures it must be at a temperature greater than 0° C up to about 10^{7
} Pa. At about 2x10^{8} Pa, about where I
have estimated the pressure to be 120 km down, it could remain liquid down to about -50° C. If there is liquid water, there would have to be some heating mechanism—vulcanism, radioactivity, tidal
force friction, etc . I don't believe the pressure itself would cause any heat and certainly not fusion.

QUESTION:
The no-communication theorem seems to forbid faster than light communication between entangled things (particles seems to specific here). 'Setting' the quantum state destroys the entanglement.
There are however now ways to "gently measure" the superposition state without destroying it.
This would appear to permit a scheme where faster than light communication is possible without violating the "no-communication" theorem.
Take an entangled pair and send one to the moon, use gentle measurement to verify the particle on the moon is still in superposition, and measure the one on earth. The timing of the collapse of the 'moon' particle can contain information. We don't care what the collapsed state is, just the timing of it. With several particles we can use a 'Morse' code to send long and short intervals between superposition collapse.
Is this an experiment worth doing?

ANSWER:
DISCLAIMER: I am by no means no expert in the subtleties of entanglement and had never heard of "gentle measurements" until receiving this question. I have done some
quick researching on this and my answer is a quick impression and should not be judged as a fact!

I read a brief
description
of the experiment which was done. "The "gentle measurement" was performed on a single
superconducting qubit, not on a particle entangled with another. It
seems unlikely to me that this method could be applied to an
entangled pair of particles without destroying the entanglement.

QUESTION:
How we can calculate the number of photons comming out of a light bulb? Is there any equation?
Can we use the same equation with x-ray to know the total number of photons per x-ray?

ANSWER:
If a lightbulb were monochromatic it
would be pretty easy to estimate this; the reason is
that white light has a whole spectrum of different
wavelengths and the energy of a photon of a particular
color depends on the wavelength. Another issue is that, depending on the kind of light bulb you use, the re will be a great many photons not in the visible range;
if it is a light bulb
you are interested in, you are probably mostly interested in the visible light. The way that light is
measured is very complicated, even convoluted at times.
In an
earlier answer I have written one of my longest ever
answers, addressing how light intensity units are
defined; near the end of that answer I estimate the flux of visible light of sunlight to be roughly 4.4x10^{12} (photons/s/m^{2} ).
This may serve as an order-of-magnitude rough estimate of the number you seek because the amount of sunlight falling on one square meter would probably be
comparable to the light from a 100 W light bulb. The number of photons would depend on the type of bulb. An incandescent bulb has many photons outside visible whereas
an LED bulb has many fewer. This can be seen in the figure showing spectra of LED, incandescent, and CFL bulbs compared to sunlight at earth's surface. Note that
the LED has very little of its spectrum outside the visible region whereas the intensity of an incandescent bulb is still rising at the red end meaning much of the radiation
is in the form of heat.

Regarding x-radiation, at
least for medical x-rays, you could probably do a reasonable estimate for a particular machine because the dominant
parts of the spectrum are discrete peaks as shown in the second figure.

QUESTION:
My understanding is Uranium and its daughter elements naturally shed mass as they change from one element to the next. Is this change effecting the gravity of the earth no matter how slight?

ANSWER:
Lots of nuclear reactions result in a loss of mass. But, mass being a form of energy, the released energy is not lost but exists in forms other than mass. It could be heat, light,
kinetic energy of ejected particles, etc . Most of this energy does not leave the earth, so the total energy density (which determines the gravity, not just mass) of the earth is essentially unchanged.
Indeed, energy lost from the earth, some photons for example, will result in a tiny (and I mean unmeasurably tiny) change in the earth's gravity.

QUESTION:
Let' s say you're driving a car 70 mph with all the windows closed. There's a fly buzzing around your head, and then you hit a telephone pole. Obviously we know that everything in the car will hit the windshield due to the momentum, but what happens to the fly? Does he hit the windshield too or does he remain in the middle in the air?

ANSWER:
Let's talk about what happens to the air. All the molecules are moving around with speeds mostly much more than than the speed of the car. When the car hits, there is a tendency
for the molecules move forward but they don't all smash into the windshield; rather, there is a tendency for the pressure in air to become very slightly higher closer to the front of the car. The fly does not witness any appreciable difference in the air,
which is the environment he encounters, and will fly
around happily.

Actually, a more
interesting question is what if there is a helium
balloon in the car. It will move to the backwards instead of
forward, experiencing a buoyant force backwards because of the
gradient of air pressure.

QUESTION:
You're standing 3 feet from a mirror looking at yourself. Perceptually, you're seeing your body 6 feet away (3 feet to the mirror and the reflection is another 3 feet), but how far is your actual vision traveling? If you look at the mirror frame, that image is 3 feet away, but when you're looking at your reflection, does the vision travel 6 feet because the image has to travel to the mirror 3 feet and then to your eye 3 feet? Another way to ask is if you shine a flashlight into the mirror from 3ft, then the light has to travel 3 feet to the mirror and 3 ft back to your eye, so 6'.. is this correct?

ANSWER:
This is a matter of semantics. It actually makes no sense to talk about
how far your vision travels . Your vision doesn't travel. This is a virtual image so it is not
actually where you would deduce it to be. If you want to know how the light travels, that is 6 ft. If you want to know where the image is, that is also 6 ft.

QUESTION:
Two spacecrafts tethered together; one heavier than the other tumbling through space, where the heavier one gets an intermittent push, and its momentum of no more than 10 N per day over a year; what would its maximum velocity be? Or would it keep on accelerating after no further energy is applied to the system?Two spacecrafts tethered together; one heavier than the other tumbling through space, where the heavier one gets an intermittent push, and its momentum of no more than 10 N per day over a year; what would its maximum velocity be? Or would it keep on accelerating after no further energy is applied to the system?

ANSWER:
This question is problematical on several levels. First, by "tumbling through space" I presume that you mean that they are rotating about the center of mass of the pair with the tether
taut. In that case, the system will have both translational and rotational energy and momentum. You have specified that the force is applied to the heavier one, not the center of mass of the system,
so the result will be that part of the enegy you add will change the translation energy (½(m+M )v ^{2} ) and part to the rotational energy (½Iω ^{2} ). Similarly for linear
and angular momenta. When no forces are being applied, the system will have some constant energy shared by the two. But if you continue adding energy you will certainly continue speeding the pair up without limit
for a long while; eventually you will have to treat the system relativistically when speeds become comparable to the speed of light. What will absolutely not happen is that
it would "…keep
on accelerating after no further energy is applied to
the system…" This is simply Newton's first law. If you want to get more detailed, you would have to specify not only the only force but the time during which it was applied (force times time=impulse) and work done (force times distance applied).

QUESTION:
In several online references to earthquake magnitudes,
e.g
Wikipedia states that the energy released is proportional to the 3/2 power of the shaking amplitude. I'm puzzled by this, as in all other wave phenomena I know about e.g Alternating Current, light, sound etc the power is proportional to the square of the wave amplitude?

ANSWER:
In all the examples you cite the waves are monochormatic, they have a single
wave length and frequency. But, an earthquake is a spectrum of many frequencies and even defining what you mean by amplitude is tricky. I am not very well versed in geophysics, but I would
assume that the 3/2 is not derived from first principles but rather is approximately determined empirically from measurements.

QUESTION:
Assume you travel at a percentage of lightspeed and return in say one year of travel and everyone has aged say 25 years to your one year. If you listen to a radio broadcast from your point of origin what will you hear?

ANSWER:
It will be different on the way out than the way back. Although the example in a
previous answer has
`every person on earth having aged 8 years more than you, not 25, the discussion in that answer
will let you get the idea of what to expect.

QUESTION:
Can two strings connected at the top (like an upside Y design) pick up double the amount of weight (equal amount of weight at the end of each string) rather than just one string alone?

ANSWER:
The figure shows the forces I want to talk about. Focus your attention first on the knot where the three strings meet. There are three forces on the knot,
T _{1} , T _{2} , and
T _{3} . Assuming the weight
is uniform, the tensions in strings 2 and 3 are the same; I will call that tension just
T . Also shown are the components of 2 and 3,
x being horizontal and y vertical. Also, because of symmetry, the
x -components of
2 and 3 cancel each other out. So we can write that T _{1} -T _{2y} -T _{3x} =0=T _{1} -2T cosθ ,
or T =T _{1} /(2cosθ ).
Next, focus your attetion on the block which has three
forces on it, its weight W and the two strings attached
to it but in the opposite direction as they pull on the
knot (I have just drawn the components here.) Again, the
x -components cancel out and we have W -2T cosθ= 0.
Putting both results together we find that T _{1} =W ,
the same result as if you were lifting it with one
string. Actually, it is much easier to look at all the
forces on the block plus the knot; that way anything
between the two is irrelevant. You will always find that
to lift something your must exert an upward force equal
to its weight.

QUESTION:
I don't do homework anymore. Not at 75 years old. But I do like learning physics.
Please explain how matter can be solid given that like charges repel and electrons make up the outer clouds (I almost said shells - showing my age) of atoms. Shouldn't atoms repel each other thus making solids impossible?

ANSWER:
Because the atom has a positively charged nucleus,
it has zero net charge. So if two atoms are not actually "touching" they will not exert forces on each other. If they get close enough to each other the electron clouds will start to
overlap and interact
with each other. The result is often that the atoms will actually attract each other. Oxygen and hydrogen actually mostly occur in pairs. You often learn, in a qualitative sense, that atoms share
their electrons with their nearest neighbors; that is not surprising because some of the electrons in one atom can be closer to the nucleus of the other atom.

QUESTION:
If entangled particles are like two coins spinning on the desk in front of me, and I slap my hand down on one of the spinning coins and then read that it landed on heads, did the other coin stop spinning instantly, without me touching it, and did it land on tails? I'm obviously a layman trying make sense of all of the various explanations I see on the internet.

ANSWER:
I like the analogy of a coin with two possible "states", heads up or tails up, to illustrate a spin ½ particle which has two possible states, spin up or spin down. And we could imagine
the spinning coin as having equal probabilities of heads or tails like an electron could have equal probabilities of up or down. Suppose that you and I each had an electron and each of
prepared the electron's state to be a mixed up/down state. Then if I measure mine and it turns out to
be down, yours doesn't instantly become up. Why not? Because these two electrons are not entangled
since they have been independently
prepared. Similarly, your two coins are not entangled and when you measure yours it could be either up or down with equal probabilities.

QUESTION:
How much electricity is required to move a 5,000 pound EV up a 20 degree incline

ANSWER:
What you really want is not "how much
electricity" but "how much energy from the battery".
Also, telling me the incline does not tell me how far up
you go, i.e ., how far up the road
do you go at a 20° incline. Suppose you go along the
road for 1 mile. Then I calculate that energy supplied if there were no friction would be 1.22x10^{7} J=3.4 kw·hr.
I have assumed that you go a constant speed. Due to
friction this would be maybe about 10% bigger.

QUESTION:
If I have two syringes connected by a 10,000 km tube filled with liquid. Syringe A is open/full. Syringe B is closed/empty. I press down syringe A causing syringe B to open/fill up. Would there be any latency between the time A is pressed, and when B fills up? There is no air in the system, and the tube isn't flexible.

ANSWER:
When you press you create a compression in the fluid at A; this compression travels at the speed (v ) of sound in the liquid to B. So the time it takes 10^{7} v .
If the fluid is water which has approximately v =1500
m/s, the time is about 1.5x10^{4} s=41.7 hours.

QUESTION:
Does the pressure of gas in a sphere follow an inverse cube law based on the principle that the change in volume is a cube of the radius?

ANSWER:
Certainly the pressure in a sphere is inversely proportional to
R ^{3} if temperature and amount of gas remain constant. But, who cares? Radius of a sphere is not what you
would call a thermodynamic variable the way volume is.

QUESTION:
My question is that if the moon doesn't have an atmosphere which means it doesn't have a pressure too how the apollo 11 astronauts spacesuit didn't explode? Based on boyle's law?

ANSWER:
What matters is the net pressure on the space suit. Although there is zero pressure outside the suit, there will be approximately atmospheric pressure P _{A} inside. The same suit on
earth could have P _{A} outside and 2P _{A} inside and not explode.

QUESTION:
Please explain how opposite processes, fission and fusion, both release energy.
It seems to me that fusion would need energy input to
fuse atoms, yet it releases energy.

ANSWER:
See an
earlier answer .

QUESTION:
I read the Rama series by Arthur C. Clark and I have a question. There is a circular 'ocean' that is held in place by the centripetal force of a revolving cylindrical spaceship. One bank of the ocean is level with the fluid the other bank is 50 ft high. A character studying the spaceship states that the 50 ft is to contain the ocean while the ship accelerates and therefor they can calculate a maximum acceleration of the ship. Is this possible? How would they do it?

ANSWER:
This is a fairly easy problem as long as you used some reasonable approximations. I will assume that the depth of the ocean (d ) is very small compared to the radius of the cylinder (R ) so
that the centripetal acceleration (a _{c} )is approximately equal at all depths, a _{c} =Rω ^{2} , where
ω is the angular velocity. I
have done a bit of research and found some pertinent information:

R =8,000 m,

The ship is 50,000 m,
long,

the "ocean" is a belt
L =10,000 m wide located at the center of the ship,

the bank is h =500 m
high (not 50 ft),

ω =0.25 rpm=0.026
s^{-1} ; therefore, Rω ^{2} =a _{c} =5.48 m/s^{2} ,
a little more than ½g .

I could find no
information about d but it needs to be small compared to
R
as noted above.

The first figure shows a schematic sketch of the ocean. I am going to focus on a small slice, indicated in red, in doing my analysis. Whatever happens here will be the same as any other slice I focus
on. The acceleration of the whole ship will be to the right; you can imagine that the water would want to be pushed toward the left and we want to ultimately want to find what the maximum acceleration is if the 500 m bank
is to contain the water.

The second figure shows my slice of the ocean when the ship has some acceleration
a to the right. In this figure, the axis of the cylinder is straight up 8,000 m above the bottom of the ocean.
It is very important to note that x - and y -axes have very different scales; I do this for clarity since angles get small. The surface of the water will be, as drawn, a
flat plane with an incline angle θ . I know this because the acceleration is uniform
in both x and y directions. If a =0 the surface
will be where the green dotted line is. I will now analyze the problem by focusing on a tiny piece of the water
of mass m on the surface; any location will work because of the uniform accelerations.
I am going to do this problem from inside the ship; as you know, in an accelerating frame of reference Newton's laws are not correct but you can force them to be true by
inventing fictitious forces; the fictitious forces here are -ma
and -ma _{c}
shown in yellow as well as the net fictitious force. Of course there is
also a real force (not drawn) on m , the force by the surrounding
water on it, and it must be equal and opposite the net fictitious
force because we have made this a statics problem. So the fact
that the net fictitious force is normal to the surface is not an accident, it is a necessity;
fluids can exert only normal forces at their surfaces, not
tangential.
Now by simple geometry we can determine the angle θ to be θ =tan^{-1} (a /a _{c} ). We can now do the case in point to find the
maximum acceleration, a _{max} =a _{c} tan(θ _{max} )=5.48x(500/5000)=0.548
m/s^{2} and θ _{max} =5.71°.
(Note, however, that slowing down is also an acceleration; in
that case a would point backwards and
the water would spill off to the right. To slow down you would
have to turn the the ship around first.)

Finally, a few comments on the above
analysis:

I have drawn the second figure so
that d=h= 500 m. Everything I did above was right if d≥h.

If d<h the water
eventually, as acceleration increases, will no longer extend
all the way to the right hand shore.

With d≥h= 500 m the surface
always goes below the a =0 plane exactly halfway between the
two shores. This is because all the water going to the left
half must leave the right half.

QUESTION:
I'm a physics student who just had a few questions
and I would really appreciate it if you could answer
them for me: I was studying granular materials and how
if you fill up a bottle with rice and then insert a
stick or spoon into it, then at a certain level of
immersion, you can pick up the whole bottle using the
spoon. I know that all of this relates to the compact
density of the granular material but what do you think
are other things I should focus on to better understand
this topic? Do you suggest that I talk about any other
physical properties of granular materials to understand
this phenomenon (like dilatancy or shear force, if so
then how are these concepts linked with the topic?)? I
also would love to know what you think are the
parameters of this experiment (of lifting up bottles of
rice by immersing a stick or a spoon to a certain
depth). What are the things that are being assumed when
explaining this phenomenon? What things will cause the
experiment to not work? Like is there anything we should
take care of (like moisture content of granular
materials or the shear force being applied to them)? Are
there any equations I should acquaint myself with?

ANSWER:
Your question violates the site ground rule requiring "...single, concise, well-focused questions..." Still, I will make some suggestions on data you should consider taking. If you are a
student, it is probably better to approach this problem empirically rather getting into theoretical analysis too deeply. I would choose one material, rice for example, and think about what variables might come
into play. What you can most easily measure is the depth at which you can first lift the container,
i.e. the upward force is equal to the weight of
the container plus grains. Some variables affecting this are the height of the container, the cross-sectional area of the container,
the shape of the container, the surface area of the "stick", the material the stick is made of, the shape of the stick, etc . And I would keep it simple, cylindrical containers, sticks with only vertical surfaces (circular or square
or rectangular crossections). I would try to use
containers which were much lighter than the granular
material. This would give you a lot of data and graphing it might yield interesting results. You might now try the same process with a different matrial,
e.g. dry sand.
To tell the truth, I really do not know much at all about this kind of physics but I do know that analysis of these kinds of problems usually involves computer simulations, not first-principle physics.
In fact I cannot actually envision being able to lift a container of rice like this; perhaps using a bottle, as you have done, with a neck is required and my suggestion of an open topped cylinder
is impossible. Good luck!

QUESTION:
I have a question about gravity
So as we know gravity is significantly less on high mountains or tall buildings and increases as we lose height. My question is that if we have 2 objects with the same volume and mass, 1 falls on a mountain and one falls on a beach ( note that they are released from the same height based on the surface they shall fall on )
Will they both reach the ground at the same time or will their speeds be different?

ANSWER:
If you neglect air drag, the one on the mountain will experience a
smaller weight force and will therefore have a smaller acceleration. It will therefore
arrive at the ground later and with a smaller velocity. But air drag might very well not be negligible. On the mountain top the air density is less dense than at
sea level and therefore the air
drag will be smaller. Now it is a more ccomplicated problem.
If, in addition to your stipulation of equal masses and volumes we stipulate the same shapes, the time it would take to reach the bottom
would depend on the height of the mountain and the height of the fall.
(I have assumed that the height of the fall is small
enough that one can approximate that the change of air
density and the gravitational force are negligible.)

QUESTION:
If a fly was the bottom of the cylinder that weights 100 n. And then the fly flies around the open cylinder what is the weight of the clyinder is less than 100 or equal to 100n

ANSWER: Also important is the weight of the air in the cylinder. The weight of the cylinder is always equal to 100 N; the question is: if the cylinder (and fly at rest) are sitting on
a scale, what does the scale read when the fly flies? See an
earlier answer .

chis question has no answer because heat is defined as the flow of energy, not an amount of energy. For a detailed discussion of thermodynamic terminolgy, see an
earlier answer .

QUESTION:
It seems evident that global warming is driving more moisture into the atmosphere. Is it possible that moisture is escaping to low-earth orbit, never to return?

ANSWER:
The velocity necessary to achieve near-earth orbit is about 18,000 mph. The most probable speed of a water molecule at 120°F is about 1220 mph. Escape velocity from the earth is about
25,000 mph. I don't think you need to worry too much about water molecules escaping from earth.

QUESTION:
I've seen that a recent AI analysis of particle collider data suggests that a proton has an extra charm and anti charm quirks, if this analysis proves to be a discovery above sigma 5, could it help explain dark matter?
I just haven't seen anyone talk about this, thanks in advance

ANSWER:
Why would you think this would have anything to do with dark matter? The mass of the proton is known to high accuracy so these data would only alter the model for the internal
structure of a proton, not its mass. No change in mass means no solution to the unknown extra mass supposedly implying dark matter.

QUESTION:
let's say a bag of rice was dumped from a flying plane, and you knew every conditions that would act on each grain. Every gust of wind, complete topographical information, the position of each grain of rice, exact gravity force, everything. If you had unlimited resources, computing power, time etc. Would it be able to calculate where every grain of rice would end up?

ANSWER:
These are some pretty severe suppositions you are placing. And you did not include effects on the grains due to their shapes. Let's start with a single grain. In principle this is a classical mechanics
problem, solvable. But there are some serious problems with what will surely be a major contributor to the problem, air drag; there are quite good approximations for the air drag force, but they are not exact.
For that reason, you could only predict approximately where any particular grain would land. If there are many grains, there will be interactions among them. These interactions would be mainly due to disruptions in the air
flow due to the motion of the grains themselves; and if two grains collide, it would be virtually impossible to predict the velocity of each after the collision. Maybe most important is that the motion of the air is chaotic,
not predictable for any particular position at any predictable time. I could envision the positions being predictable if the experiment were performed in a perfect vacuum and care was taken in dumping them
that they be far enough apart that they never suffer a collision.

QUESTION:
If I were in London England and launched a drone with 6 hours of battery and set it to hoover perfectly still, when the battery died would the drone land in London or America?

ANSWER:
What you need to understand is that your drone is rotating around the earth's axis just like anything "at rest" is. When you set it up to hover over London it is already moving right
along with London. The drone flies relative to the air, so if you set it up when the air is still
and if later a wind starts blowing, the drone will move with the wind; but it will certainly not get blown as far
as America is from London in six hours.

QUESTION:
We're having a lively debate at work regarding best driving practices for fuel economy, ad we've finally settled into two camps on one specific topic; what to do when approaching a hill. Group 1 believes that the driver should always coast on downslopes and never touch the gas. Their reasoning being to cash in on all the free momentum you can get. The 2nd group of us dissent and claim that's exactly when you should accelerate, with the rationale of using gravity as a catalyst for a more fuel efficient fuel burn. Is there a clear cut answer to this; and if not, what's a good way to think about the problem?

ANSWER:
As you go down the hill do you suppose you use more fuel when coasting or accelerating? It seems to me the answer is obvious. Suppose the hill is 1 mile long and the accelerating car consumes
0.1 gallons of gas and the coasting car consumes 0.05 gallons; for that leg of your trip the coasting has 20 mpg and the accelerating has 10 mpg. If your car is a hybrid you will consume no gas and the energy
you gain by speeding up will be saved in the batteries so this mile was free when computing mpg.
I'm afraid that "using
gravity as a catalyst for a more fuel efficient fuel
burn" means nothing to me.

QUESTION:
When radiation passes through a GM-tube the gas inside is ionised. How come that a GM-tube is reusable if the atoms in the gas change. They lose electrons, so those atoms wouldn't be able to absorb the energy of the radiation the same as gas that hasn't already been ionised. So wouldn't the gas that can be ionised eventually run out?

ANSWER:
These tubes are filled with a noble gas, helium, neon, argon, etc .
Many atoms (let's say N) become ionized as the radiation
passes through the gas. But they don't stay ionized
because the we now have N free electrons and N positive
ions and they recombine because the Coulomb force
attracts electrons to to the ions.

QUESTION:
I am out of high school and looking to strengthen my understanding of physics, this question has been bugging me. For third law, it states that every action force has an equal or opposite reaction. How does third law apply to rockets? Everyone tells me that it is the act of throwing something or the opposite reaction force that propels the rocket or the balloon, but I think they are all wrong it is not because of the releasing of the fluid, I think it is the net force within the system of the rocket ship or the balloon that pushes it up. Am I correct?

ANSWER:
I suggest that you read a
recent answer on jet ski propulsion. This addresses the issues about
rocket propulsion. Your explanation for rocket propulsion is wrong. Newton's third law is indeed the correct explanation.

FOLLOWUP QUESTION:
So I read the dialogue/conversation(chat?) about the seadoo and the "friend" and yes you are completely right (obviously) about how the seadoo would still acccelerate even on air, it's just a different medium. The breakdown of the engine makes complete sense to me aswell, and 100% third law applies, from the impeller being attached to the boat or the body of the vessel and the impeler is forcing out water to the nozzel, so impeller pushe water out water has interaction with impeller impeller is pushed by water in opposite direction voila movement. But from the little information I gathered, it seems rocket engines are different. They are like mini explosions that are contained, and I understand third law is still applied when in earth atmosphere and stratosphere the air will cause drag from it interacting with the vessel but third law doesn't make sense in space which has no air ( their are particles in space like neutrinos traveling understandablly but rockets don't push off of that light sails do.) the point is with the seadoo it is interacting with the water medium and in the other example the air medium to traverse. But in space I think third law does not apply, due to their being a vacuum.
But yeah on earth it makes sense, the thrust has to overcome the
force air has on the rocket or the drag, so Newtons third
applies to earth but I don't think for space.

ANSWER:
A jet engine draws air from the atmosphere and adds kinetic energy (speeds it up using a fan driven
by fuel carried in the airplane) to the air which it ejects out the back; the reason that this will not
work in space is that there is no air to draw in, NOT because the jet needs air to "push against".
The same principle applies to the jet ski motor but water is the propellant, not air;
the air has nothing to do with why the jet ski works outside of the water (be sure to note that
the
question of the 'friend' states "…with a source of
incoming water for the water pump inside it…").
It would work just as well in space as long as you fed it water; similarly, a jet engine would work just fine in space as long as you provided it with air.
A rocket engine works on exactly the same principle as a jet engine except that the fuel itself is both the impeller and the impelled. Something gives kinetic energy to the molecules (your little explosions); that
something is chemistry. But each molecule which is going faster than it originally was, has experienced a force accelerating it out the nozzle and therefore the exiting molecule exerts an equal force back on
the rocket ship.

Alas, I believe that one of the most used and least understood principles in physics is action/reaction to describe Newton's third law. Those words should be discarded and the statement of the law should be "If object A exerts a force on object B, object B exerts an equal and opposite force on object A".

QUESTION:
My question is, why is it that fission occurs when a neutron and a 235Uranium collide?

ANSWER:
Any decay mode of a nucleus which can occur (is not forbidden by conservation rules and is exothermic) will happen with some probability. Many nuclei are unstable to β decay and many
heavy nuclei can decay by α decay. α decay is just a very asymetric fission. In fact, ^{235} U undergoes spontaneous fission without any neutrons at all, but its half life is about
3.5x10^{17} years; so the rate of spontaneous fission is about 14 fissions/gram/day. However, if you can find a way to add energy to the nucleus you can greatly increase the probability for
fission to occur. If you want to have a picture of
what is happening we can use a semiclassical model of the nucleus which is very successful, the liquid drop model. ^{235} U is not spherical but shaped
like an American football (prolate); when you strike it with a neutron or the neutron is absorbed, the resulting nucleus is highly excited which means, for a prolate liquid drop, that it is simultaneously rotating and
vibrating. If it rotates fast enough the centrifugal force stretches it which, is sufficiently stretched, causes a neck to form which then makes fission likely. See the figure for a cartoon of the fission.
The liquid drop model was developed by Aage Bohr, James
Rainwater, and Ben Mottelson who received a Nobel Prize
for their work.

QUESTION:
If I measure my velocity in relation to light, can I determine if I am stationary or in motion? Since light is measured as the same speed (in a vacuum) by all observers? I.e. I can calculate my speed as x% of the speed of light letting me know if my "spaceship" is stationary or moving with respect to that light?

ANSWER:
I cannot understand how you can measure your speed relative to something which will always be seen by you as moving at the same speed regardless of your speed. No matter
what your speed is relative to the some other object (e.g. , like the ground if you are in a car), the light you see will always have the same speed relative to you.

QUESTION:
Since acceleration is a change in speed or direction,
And moving in a circle at constant speed is therefore accelerating;
And since continued acceleration requires an external force,
Isn't any point on a spinning object in space accelerating forever without an external force ?
For example, I understand that accelerating a rocket in space to v, the rocket will continue at v until it encounters a large object.
But spinning the rocket and then letting it go, what keeps the rotation going indefinitely, since any point on the rocket is accelerating due to the spin ?

ANSWER:
For your first question, let's consider you standing on the spinning earth. Are there any forces on you? Of course there are—the gravitational force (your weight) W ,
the normal force N of the ground, and a frictional force f with the ground, as shown on the figure.
So you, standing somewhere in the US northwest, experience these three forces
drawn in red; I have also drawn the components parallel
and perpendicular to the earth's rotation axis. You are also moving in the circle shown in yellow. So the force providing the centripetal force is the vector sum of all three components perpendicular
to the axis.

For the second part of your question, why does the rocket moving with velocity
v continue moving with the same velocity if there are
no external forces on it? Because of Newton's first law which may be stated that the linear momentum (mass times velocity) is conserved (remains constant) if there are no external
forces . If the rocket is
spinning, why does it continue to spin? Also because of Newton's first law which, in this case, may be stated that the angular momentum (moment of inertia times angular velocity) is conserved (remains constant) if
there are no external torques . And the force which
causes each point to continually accelerate is just the force
which holds the whole rocket together, electrical forces between
the atoms.

QUESTION:
I have thought
about F=ma and p=mv but I am missing something and it's
driving me crazy. This actually has a practical
application for people like me. If you skydive (90 kg
person), you accelerate at 9.8 m/s^{2} for
approximately 3-5 seconds (estimate) before you reach
terminal velocity of about 120 mph (54 m/s). The force
required for slowing down is the deceleration that
occurs when you open your chute. (At least that is how I
think of the issue). So the amount of force to slow down
to a velocity of 8 mph is much greater while in free
fall than the force required to slow down to a velocity
of 8 mph (3.6 m/s) while at a terminal velocity of 120 mph. Right?

ANSWER:
I think your problem is that you are assuming that there is some constant force acting which slows you down from 120 to 8 mph after the chute opens. In fact,
the force which slows you down and causes you to reach some terminal velocity (120 mph without open chute, 8 mph with) is air drag force which depends on both the object's speed and its cross sectional area
A . A good
approximation of the drag force near sea level is ¼Av ^{2} ; this approximation is only true if you work in SI units (metric) which is why I converted your speeds to m/s in your question.
So, when you jump out of the plane there is no drag but as you accelerate down because of your weight
(mass times g , mg ) you go faster and the drag force will get bigger and bigger until it is approximately equal to weight down and you will stop
accelerating, ¼Av ^{2} =mg . This lets you write an expression for the terminal velocity, v =√(4mg /A ).
With your numbers, 54^{2} =4x9.8x90/A so the effective
area is 1.2 m^{2} . Now the chute opens and the
area gets much bigger which means that you are now going
much faster than the terminal velocity so you will slow
down; 3.6^{2} =4x9.8x90/A or A =272
m^{2} . Before you open your chute you can speed
up by orienting your body straight down (smaller area)
and slow down by falling spread-eagle.

QUESTION:
When fusion or fission happens, the nuclear particles lose mass, how?

ANSWER:
If you consider one nucleus for fission or two for fusion and imagine them as each isolated systems, the total energy of each isolated system must remain the same. Don't forget, mass is energy, E=mc ^{2.} So, if mass is lost, as you point out, some energy of the system is now missing. If that is the case,
then the missing mass energy shows up as kinetic energy of the final parts of the system.

QUESTION:
Does a
'sea doo' have to be sitting in water in order for it to accelerate? This ignores the fact that it needs a water intake. In other words, if the sea doo was suspended over the water, with a source of incoming water for the water pump inside it, would it still accelerate through the air, without sitting in the lake, and having the outgoing water stream hitting the surrounding water in the lake, in order to provide forward movement at the usual speed?

ANSWER:
Hmm...are you the 'friend' whose question I
recently answered?!
If so, are you angry to be wrong? The answer to your question is that the sea doo would definitely accelerate; it is simply a jet engine which uses water rather than air for propulsion. If you read carefully my
earlier answer , you will see that
the propulsion is not due to the ejected water pushing on the water outside but rather due to the ejected water pushing backwards on the impeller blade and therefore on the whole craft.

QUESTION:
I am curious as to what happens when you crack a whip, but specifically regarding how much heat is liberated when it reaches supersonic speed. Let's say for example that I crack a whip while surrounded by a highly flammable substance in the air. Would the heat released be enough to ignite this substance and create a fireball?

ANSWER:
The end of the whip has a very small mass and therefore its kinetic energy is pretty small even though it is moving very fast. The little research I have done leads me to suspect most of the energy goes into the shock wave, not heat.
If you want to get a lot of details about the physics of bull whips, go to this
link and scroll down to
April Jennifer Choi ; be
sure to watch the video there too. I have included a
high-speed camera picture of the shock wave.

QUESTION:
I have been having
an argument with a 'friend' for 3 years now about how
seadoo thrust works. He says that the water must hit
something in order to move the vessel forward. I believe
that it is basic mass * velocity = force.

ANSWER:
Your friend is wrong. I would guess
that his opinion stems from that old issue about how a
rocket works: the exiting hot gas must push against the
atmosphere to move the rocket. Therefore rockets do not
work in empty space, right? WRONG! Of course, you are
completely wrong when you say that the force is equal to
the product of the mass and the velocity. Mass (kg)
times velocity (m/s) does not even have the dimensions
of force (kg·m/s^{2} ). The figure shows a
diagram of the jet. The engine is not in the figure but
is spinning the drive shaft 8. Attached to the shaft is
a propeller, referred to as the impeller 10. As it spins
it exerts a force on the water which pushes water out
the nozzle in a stream B. The water pushed out is now
replaced by drawing water in from the outside through a
hole in the bottom of the boat (on the left in the
diagram). Now comes the physics of the whole thing:
Newton's third law says that if the impeller exerts a
force on the water, the water exerts an equal and
opposite force on the impeller. But the impeller is
attached to the boat, so the force on the impeller is a
force on the boat which pushes it to the left. (I have
deleted the second question about the water and the wall
because it makes no sense; resistance is not
defined.)

QUESTION:
Spinning metal shafts power our world as in generators, automobiles, etc. My question is does the very atomic or molecular center of a shaft also spin or does it remain still.

ANSWER:
First of all, you cannot apply the concepts of classical mechanics at a microscopic (atomic) level. The notion of the exact point of the axis of rotation being remaining
at rest assumes that the rotating object
has a uniform mass distribution; at the atomic level it
is not uniform at all. Also, in classical mechanics you
can talk about a point which occupies zero volume and
has no rotational degree of freedom and an atom
satisfies neither of those conditions.

QUESTION:
Re: speed of gravitational waves. Why is it said to be the same as the speed of light? The speed of light =1/root(eu) where e is the vacuum permittivity and u is the vacuum permeability. e is an electric field parameter and u is a magnetic field parameter, both of which affect photons.
Why would electric and magnetic parameters affect the speed of gravitational waves? It doesn't make any sense.

ANSWER:
W e do not have a theory of quantum gravity, and therefore the graviton, which would play the
same role for gravity as photons play for electromagnetism
("messenger" of the field), is
merely a speculation. However, it is believed that a
graviton would have to be massless and therefore would
travel at the speed of light. Actually, there is a quite
good measurement of the speed of gravity which comes
from the gravity wave measurements at the LIGO/Virgo
gravitational wave detectors. Most events detecting
gravitational waves are from collisions and merging of
two black holes, and therefore we do not detect any
electromagnetic radiation since light cannot escape a
black hole. Rarely, though, gravitational waves from a
collision of two neutron stars is observed. In 2017 such
a collision was observed and 1.7 s later a gamma ray
burst was observed; this implies that the speeds of
light and gravity differ by less than one part in 10^{15} !
The 1.7 s difference can be attributed to the fact that
photons are affected by interaction with the
interstellar medium (not a perfect vacuum) thus slowing
them ever so slightly.

FOLLOWUP QUESTION:
The Physicist said: "However, it is believed that a graviton would have to be massless and therefore would travel at the speed of light."
this is what I don't get. why would EM parameters affect the speed of gravity. Here is my suggestion:
Early after the big bang, the four forces are thought to have been united. I suspect that when gravity and EM were united, they shared a parameter that determined the speed of all of their waves. i.e., they were the same, 300,000 km/s.
So speeds of grav waves and EM waves share a common origin and it is not correct to say gave waves are governed by EM.

ANSWER:
Since I state on the site that I do not do astronomy/astrophysics/cosmology, I will not comment on your hypothesis about the big bang and unified forces at the time.
However, gravity and electromagnetism are already linked and I will briefly explain how.
The theory of special relativity had its origin purely
in electromagnetism—the speed of light is
independent of the motion of the source or the observer.
And, although Einstein stated it as a postulate, that
fact is already a consequence of the transformation
properties of the waves which emerge from Maxwell's
equations, perhaps the best understood of all field
theories. Having formulated special relativity, Einstein
next turned to extend it to have a broader application,
the theory of general relativity. Of course, general
relativity is the best theory we have of gravity, and
since it is evolved from special relativity which
evolved from electromagnetism, we can say that gravity
has its roots in electromagnetic theory. Gravity must
then obey all the findings of special relativity, in
particular that the speed of anything must not exceed
the speed of light and a massless particle must travel
at the speed of light.

QUESTION:
leaving aside for the moment what relativity theopry says, do you know of any experiment that has been performed to show that pure energy can gravitate?

ANSWER:
What is "pure energy"? This has no meaning in physics,
energy is energy. And by "gravitate" do you mean responds to a gravitational field? So, I will choose light as
your "pure energy". It is an experimental
fact that light is bent when passing through a gravitational field, so it "gravitates".

QUESTION:
my book says, "friction helps in transferring the motion from one body to another". I am not getting this. can you please explain this.

ANSWER:
Here is just one example of many I might think of. There are two wheels side by side; one is rotating with some speed as shown in the upper half of the figure, the other is not rotating. We
now move the rotating wheel so that it eventually touches the unrotating wheel. If there were no friction between the two wheels, the rotating wheel would keep rotating and the unrotating wheel would
stay stationary. But if there is friction, the stationary
wheel would feel a torque due the frictional force from the rotating wheel. This torque would cause the the stationary wheel to speed up; Newton's third law would
tell you that the rotating wheel would feel a frictional force in the opposite direction which would cause it to slow down. Some of the motion of one wheel has been transferred to the other by friction.
If the two wheels were identical, they would eventually
stop slipping would and the wheels would rotate with the
same speeds but in opposite directions.

QUESTION:
Is alternating
current the same principle as reverse polarity?

ANSWER:
Neither one of these is a "principle" but they are related to each other. Alternating current refers to a current which is changing direction; to change the direction you need to change the polarity of the electric
potential which is driving the current.

QUESTION:
William Herschel's
rainbow experiment discovered infrared rays. Looking at
the results of the experiment, we see infrared rays
hotter than visible red light and red hotter than blue
light. Why? The wavelength of infrared is longer than
red light and the wavelength of red is longer than blue.
Additionally astronomers tell us red stars are cooler
than blue ones.

ANSWER:
A single photon has an energy proportional to its frequency and therefore inversely proportional to its wavelength. So blue photons have more energy than red photons;
this explains why red stars are cooler than blue stars.
So your question is why Herschel found higher temperatures at the red end of the spectrum than the blue.
The main explanation was, because he used a prism to
create the spectrum of sunlight, and the prism has an
index of refraction which is a function of the
wavelength. That is what causes different colors to bend
by different angles. But it is not a linear function of
wavelength, rather the curve shown in the figure. Note
that the short-wavelength (blue) index of refraction
changes much more rapidly than at long-wavelengths
(red). Therefore the blue portion of the spectrum is
much more spread out than the red and therefore the
thermometer for blue measures a smaller fraction of the
total blue spectrum than the thermometer for read
measures for the red part.

QUESTION:
I have seen many years ago a nasa invented patent which i will present to you in a link, which was supposed to propel itself by using double pendulum method. Im asking of you to check it, whether its false or true.
https://youtu.be/6A_cEZCXBIc I hired a specialist in blender to make this animation, based on my memory, because i couldnt find it.

ANSWER:
Sorry, I can make no sense of this. What I can tell you is that no isolated object can propel itself. If there are no external forces on an object its linear momentum (mass times velocity) must remain constant.

QUESTION:
Say I had an extremely bright flashlight, one impossibly bright for our current technology, and I shone it into space. If I were to then move that light a bit, would I be able to see, in real time, the ray of light bend as it catches up? Or would it be instant? This is specifically referring to the stratosphere

ANSWER:
Brightness has nothing to do with it—since
there is air and dust up through the stratosphere there
would be some scattering allowing you to see the beam
for any intensity. The question is whether you could see
a bend. The stratosphere is approximately between 10 km
and 30 km. How long does it take the light to travel 30
km? t =3x10^{4} /3x10^{8} =10^{-4} s.
Now, suppose you move the light through θ =10°=0.175
rad in T =0.1 s; if the transmission were
instantaneous the beam at R =30 km would move with a speed
of V =3x10^{4} x0.175/0.1=5.25x10^{4
} m/s so it would go a distance of D =5.25x10^{3}
m. The light beam is not (as you state) a straight line
but lags the straight line a bit. All this is shown in
the figure where blue is for straight line travel
(instantaneous) and red for actual (but not to scale as
we shall see!) path. But the actual beam at 30 km will
lag the instantenous beam by t =10^{-4} s which means that the actual distance
traveled in 0.1 s is D-d=V(T-t )=5.25x10^{4} (0.1-10^{-4} )=5.25x(10^{3} -1)=5.245x10^{3} .
So (D-d)/D=0.999; so d is far smaller than
sketched in the figure, only 0.1% of D . I suggest that it would
be virtually impossible to visually detect any curvature
of the beam.

QUESTION:
Why do we always see the same stars in the sky as the ancients did? As if we are ment to be traveling through space... wouldn't we see different stars and certainly wouldn't Polaris at least not be the same centre point?

ANSWER:
It mostly has to do with the enormous distances to stars from earth. I looked up some data: the speed of a typical star in our galaxy is about 8x10^{5} km/hr which is about 10^{-3} c where
c is the speed of light and the radius of the
Milky Way galaxy is about 5x10^{4} ly (light
years). So I will choose as a likely distance to a star
in our sky 1000 ly and a reasonable speed perpendicular
to our line of sight of 10^{-3} c . In
one year the star will move 0.001 ly. I
have sketched this in the figure (obviously not drawn to
scale); the angle θ is 10^{-6}
radians, about 0.0001°. In 1000 years it will only
have moved about 0.1°, measureable but not even
noticeable. So to see significant changes to
constellations you really need to wait hundreds of
thousands of years. Also of interest is that the earth
has a precession of its axis with a period of about
26,000 years. This does not result in any change of
relative positions of stars but rather moves them
around; in about 12,000 years Vega will be the north star, not
Polaris. Polaris (and everything else) would have moved about 40°
in the sky.

QUESTION:
I have a very mundane question. I have a job pushing wheelchairs up driveways and ramps, and also walking wheelchairs down ramps and driveways. So I'm interested in the force needed to do that based on the incline of the ramp or driveway.
Assume that a wheelchair has solid rubber tires that don't deform much. Assume also that the combined weight of the person and wheelchair is 400 pounds. How much force would be needed to push the person and chair

On a flat surface with no incline

Up an incline of 5 degrees

Up an incline of 10 degrees

Any other combinations of weight and incline that you might find interesting or instructive.

Also, would the force going in the reverse direction (walking the wheelchair down the driveway instead of up) be the same.

ANSWER:
I will solve the problem in general, then I can apply that to all your questions. In the diagram I show all the forces on the wheelchair plus occupant:

W , the net weight; I also show the components of W along and normal to the incline, respectively;

N the normal force which the incline exerts on the chair; this is the force which the incline exerts to keep the chair from falling through it.

f , the frictional force impeding the motion; this can be approximated as f=μN where μ is a constant. After searching around I found that
μ can be approximated by a small number of about 0.02.

P is the force which the lady (that's you!) pushes to keep the wheelchair moving with a constant speed.

All forces must add up to zero; we achieve this summing separately x and y forces:

N-W_{y} =0=N-W cosθ or N=W cosθ , and

P-W_{x} -f =0=P-W sinθ-Wμ cosθ, or P =W (sinθ+μ cosθ ).

Now I can answer your questions:

No incline means θ =0, so P =400x0.02=8 lb.

For θ =5, P =400(0.0872+0.0199)=42.8 lb.

For θ =10, P =400(0.174+0.0197)=77.5 lb.

Note in #5 that going down the incline the friction helps you; going up you have to push against it.

For going down the incline, θ is negative, so the equation for P (which is now a pull rather than a push) is
P =W (-sin|θ|+μ cos|θ| ). So for θ =5, P =400(-0.0872+0.0199)=-26.9 lb.

QUESTION:
I understand why the twin paradox works in theory, but if man had the means to produce all the requisite energy to propel a spaceship at any acceleration for any duration of time , could a twin physically leave earth in a space craft and come back 50 years later noticeably different in age from his/her twin? Or would the amount of acceleration required to obtain the necessary relative speeds be to much for a human to bear?

ANSWER:
You are certainly right, the usual discussion of the twin paradox assumes that all accelerations are essentially
instantenous which would make the necessary force infinite. Because we evolved in an environment where the gravitational acceleration is about
g =10 m/s^{2} we cannot endure
accelerations much larger than that. The largest a normal healthy person can endure is about 9 times
g and that only for a few seconds before blacking out.
So if you could maintain a constant acceleration of
g would be ideal, but slow; just to get to c /2,
for example (in the traveling twin's frame) would take
about 6 months. You might be interested in this
web page.

QUESTION:
I have to settle a friendly bet. My friend says if gravity was ever instantly nullified for the entire planet Earth, that people and anything not bolted down would instantly begin to fly off into space. I disagree, because there are other forces at work like
inertia, friction, air pressure, etc that I believe would need to be overcome before everything simply went into freefall. I realize that what he describes would eventually happen as those forces reconcile naturally...but how quickly would that happen?

ANSWER:
Sorry, your friend wins this bet. Everyone on earth is moving in a circle as the earth rotates on its axis.
We are in uniform circular motion which requires a force
toward the axis to keep moving in the circle. This force
is the component of your weight toward the axis of
rotation, shown as W _{r} in my figure.
If your weight W disappears, so does W _{r}
and you will continue moving in a straight line in an
easterly direction tangential to the surface. Being
tangential to the surface, you would be rising as you
moved; but the earth would be rotating beneath you so
you would perceive your path as approximately straight
up. Since the air is moving essentially with the
rotating earth, it would not much impede your departure.
Inertia is not a force. Friction from what? You are
essentially moving with the air, so no effects due to
air drag or air pressure would retard your departure.

QUESTION:
In Einstein's famous
equation E= MC^{2} , has anyone ever demonstrated the
reversibility of the equation; I.e M= E/C^{2} ; the
transmutation of pure energy into newly created matter? I'm a scientist and this simple question has stumped both myself and my peers.

ANSWER:
"Pure energy" has no meaning. I can give you a couple of examples of energy being converted into mass. The first is called
pair production . A photon (a quantum of light, electromagnetic energy)
has no mass but has an energy E=hf where h is Planck's constant and
f is the frequency of the photon. If the energy
of the photon is large enough, it can (and does) turn
into an electron and its antiparticle the positron, each
with mass m _{e} . A photon will not do
this on its own, but can be triggered to do it by an
intense electric field, for example close to a nucleus
of an atom. Another way that we can turn energy into
mass is to break apart some nucleus. A simple gedanken
is to break a nucleus into two pieces and pull them away
from each other so the two pieces end up very far apart.
Because we know that nuclei exist, those two pieces must
begin our experiment being very attracted to each other;
so to bring them very far apart from each other requires
that we do positive work (that which increases energy)
on them. If you now weigh the two pieces and compare
them with the weight of the original nucleus, you will
find more mass than you started, energy to mass.

QUESTION:
I am not a student...Brain storming. Q:in a 100' dia. X 50' high tank with a 10' dia. X 52' high cylinder attached in center,standing straight up (water tight). Fill lg. tank with water. How much pressure is exerted per square inch on inner cylinder at the bottom first few feet?

ANSWER:
The pressure in a fluid at depth d is
P=P _{a} +ρgd where P _{a} is
atmospheric pressure, g is the acceleration due to gravity, and
ρ is the density of the fluid. This will
also be the pressure on your smaller cylinder at this
depth.

QUESTION:
I'm just writing a sci-fi story and am a little pedantic about details. My question is: if my space craft is launching from the Martian surface will my occupants and their unsecured things inside the space craft experience weightlessness like we see on space stations like the ISS at some point?

ANSWER:
During the launch they will not experience "weightlessness" because there is a net force upwards to accelerate the craft and all its occupants upwards. As soon as the engines
are turned off, though, they are in free fall and will experience "weightlessness". Be sure to understand that if you define (as I do) weight to be the force of gravity on something, they are not really weightless
but only feel like they are. Once they are far away from Mars, the gravitational force will be small enough that you may say they are approximately truly weightless.

QUESTION:
Is force the only thing that can make something move? And secondly, what makes something accelerate? Is it when energy changes form?

ANSWER:
Newton's first law says that an object in motion with constant velocity will continue like that until acted upon a force; forces change the motion of an object, but no force is required
to keep it moving. Newton's second law says that forces change the motion of a particle, accelerates it.
Your last sentence does not mean anything to me.

QUESTION:
So, kinematics is the about motion, disregarding the forces, like a(t) = dv/dt. What do you call "non-kinematics", when forces are included? Would you call uniform circular motion "kinematics" if what causes the centripetal force is not specified?

ANSWER:
Kinematics is the description of motion without any reference to the cause of the motion. If you attribute Newton's second law as being the cause of any particular motion, it is called dynamics.
If you say that an object moves in a circle R
with constant speed v and has an accelleration
with constant magnitude a=v ^{2} /R
which always points toward the center of the circle,
that is kinematics.

QUESTION:
My question could be a simple questions, but still I was not able to get a proper explanation for it from the people I asked.
If an ascending balloon in the air is carrying a box and the box is released, why does the box move slighlty upward before going down?

ANSWER:
I think you will understand better if I exaggerate the situation a bit.
Instead of a balloon drifting lazily upward, the box is
being carried upward by a rocket ascending with a speed
of 1000 mph. Surely you realize that the box is moving
upward with a speed of 1000 mph. And if you now release
the box, it would not simply immediately be at rest! It
will be going upward with a speed of 1000 mph, but would
begin slowing down immediately because of the downward
force on the weight of the box.

QUESTION:
Hi, a question from a home brewer. A constant heat source is applied to a container of water. Is the rate of boil off constant or does it vary as the volume of boiling water decreases?

ANSWER:
Once all the water comes to the
boiling point, all the energy being added (heat) goes
into the water*, a certain amount energy will result in
a certain amount water evaporation. So the rate should
not depend on the volume of water.

*I assume that
radiation losses from the container and the surface of
the water are negligible.

QUESTION:
It's been proven that time passes faster on the mountain top than on the lowland. Does this speed of the passage of time continue to decrease if it is monitored below the earth's surface toward the earth's core?

ANSWER:
The stronger the gravitational field,
the more slowly time passes. If you move from the
surface of the earth toward the center, the
gravitational field decreases and becomes zero at the
center. Therefore the clock inside the volume will run
faster than one on the surface.

QUESTION:
How is sound 'carried on the wind' when it itself is a frequency?

ANSWER:
Your question does not make sense to
me. I do not see what "…itself is a frequency…"
means and what that might have to do with sound being
"…carried on the wind…" Sound is a wave
which is distubances which propogates through the air,
maybe even one particular frequency but not necessarily.
If there is no wind, you would measure the sound to have
some speed, say v_{s} . If there
happened to be a wind of speed v_{w}
and blowing in the same direction as the sound is
moving, you would measure the speed of the sound to be
v_{s} +v_{w} . So I
guess you would say that the wind is carrying the sound
at a speed faster than the sound would travel with no
wind.

QUESTION:
Can elementary particles be created or destroyed.
The impression I am under based on the confusing answers I've found regarding this question is that as far as we can tell, they cannot. Which makes sense, because my understanding of our understanding of the subject at hand is that Elementary particles (Fermions/Bosons) cannot be created from nothing or destroyed into nothing. I get that with annihilation, fusion/fission, vaporization, and plenty more I didn't list or am unaware of basically result in the conversion and recycling of particles, but that doesn't mean, to me at least, they can be destroyed.

ANSWER:
Your understanding of the words created or destroyed is not what physicists would say those words mean. You are missing
the fundamental principle here. Any closed system must have its total energy conserved, not conservation of particle kind. If two protons collide with sufficent energy
they may both disappear but not into nothing. The energy
they had (including mass energy, of course) must show up
as some kind of energy so there has been no loss of
total energy. So the collision might result in several
particles which are not protons. So I would say that the
protons have been destroyed and some other kind of
particles have been created. But you would say that the
protons have not been destroyed and the other particles
have not been created. Your definitions is too
restrictive to be useful in physics. But, using your
definitions, no particle can be either created nor
destroyed.

QUESTION:
My intuition may be wrong here...
I get the feeling that I've sat in front of electric out door heaters which are really warm, you can feel the heat on your skin, even if there is a breeze. An outdoor LPG heater I guess is warmer on a still day but on a breezy day. What's going on here? Or am I just plain mistaken?

ANSWER:
There are three ways we generally
have heat transfer:

Conduction
where heat flows through a solid.

Convection
where heat is carried by currents in fluids.

Radiation
where heat is radiated away from something hot.

In your case, only
#3 is a significant contributor to the heat you feel.
The wind has almost nothing to do with it.

QUESTION:
What is the significance of the Dipole moment in electrostatics? No one simply multiplies the distance between charges and the magnitude of the charge without a reason.

ANSWER:
What
is the significance of a monopole (a single charge) in
electrostatics? Well they exist in nature and so we
might want to understand what the electric field of a
charge will be; it turns out, as you doubtless know, to
fall off like 1/r ^{2} where r
is the distance you are from the charge. But, there are
two kinds of electric charge, positive and negative we
usually label them. And since pairs of equal electric
charges often happen in nature (for example a hydrogen
atom), we might like to know what the field of a dipole
looks like. It turns out that because there is zero net
charge in a dipole, at far distances from the hydrogen
atom you will see no field at all or at least you can
surmise that it falls off faster than 1/r ^{2} . In fact it falls off
approximately like 1/r ^{3} and it is
not spherically symmetric like a point charge field is.
If you think about it, all molecules and solids are held
together by attractive forces between electrons and
protons. So thinking of molecules requires thinking
about dipoles.

QUESTION:
If a boxer was to throw a punch wearing 12 oz gloves and then throw the same punch wearing 16 oz gloves which one would have the most impact force/damage? (...not...homework, just a debate between friends who have no one with the intelligence to settle the debate.)

ANSWER:
A little research on my part found that the only differences are the weight of the glove and the amount of padding. The 12 oz gloves are what
are normally used in matches,
the 16 oz are for training and sparring. The reason they are used in sparring is that the extra padding will prevent hurting the other guy
as much when you throw your best punch.

QUESTION:
As energy could be
converted to mass, if the einstein's equation is
applied, then the equivalent mass will be produced. Then
the neighbouring planet will gets gravitational
potential energy, so after this proces, we created even
more energy than before? If it's correct, what's the
extra energy from?

ANSWER:
I
am assuming that at the beginning of the experiment you
have a planet and some energy density hanging around
somewhere nearby. To make things simple I will assume
that the energy density of the energy hanging around is
spherical and uniform; then I find a way to convert the
energy into mass which is also spherical and uniform. So
now you are saying that the energy arising in the
planet-mass system is different from what was in the
planet-energy system because of the gravitational
interaction between the two masses. But here is what you
are missing: not only mass causes gravity, but any
energy density causes gravity. In my example there is no
difference between the energy contained in the whole
systems because all the fields are exactly what they
were before the energy to mass conversion. Believe me,
in any closed system (interacting with nothing else) the
total energy remains constant.

QUESTION:
I am a criminologist so
this is out of my area but I need this information: As I
understand it, entanglement occurs when a particle pair
is divided and then separated Such as a photon. If they
had to be together and they became entangled in would
not the maximum distance they can never be effectively
tested such a test were to be developed would be within
the speed of light distance over a given time.

ANSWER:
You have sort of hit on why Einstein called it "spooky action at a distance"! What entanglement is is two particles, two electrons or two photons, for example, which have wave functions which are mixed. (Wave functions essentially
tell you something about the particles.) The most
common experiment is to prepare the system so that the
two electrons have zero net spin. In classical physics
this could only occur if one spins clockwise and the
other counterclockwise, so these would not be
"entangled". But in quantum physics a wave function need
not be "pure" and you could get zero net spin if both
particles were 50-50 clockwise-counterclockwise and we
would say that they are entangled. Some time later, and
it could be as long as you like, you make a measurement
to determine the spin of one of the two and determine
that it is spinning counterclockwise;
instantaneously the wave function of the other
would become pure clockwise. What a measurement does it
to "collapse" the wave function and put the particle
into a "pure" state. And the catch here is the word
instantaneously means just that so any worries about
waiting for the second particle to "get the message"
that her partner has been determined to be in a pure
state are not relevant. In principle you could wait
years to make your measurement and then determine the
immediate result on the other; you would have to do some
very careful timing of the measurement of the far gone
particle but you would find that it happened at the same
time as your measurement.

QUESTION:
I have a question regarding light and mirrors. I have been taught that light consists of all colors and that things that appear to be white are only white because they reflect all light. This has me confused because there's another thing that i know of which also reflects all light, mirrors. Why is it then that mirrors create a perfect mirror image whereas a white piece of paper is only the colour white?

ANSWER: Because a mirror
has a very smooth reflective surface and therefor the
law of reflection, that a ray of light reflects at the
same angle relative to the normal with which it came in,
and therefore an image will be formed. A piece of paper
is rough and any ray of light which strikes it will
reflect out at a random angle.

QUESTION:
This question comes from my 11 year-old, who thinks about this stuff while falling asleep. Some objects, like dry wood, float because they are less dense than water. Water is heavy. Is there some depth where the weight of the water column above an object would keep it submerged, even though the object is less dense than water. If the less dense object would rise regardless of the depth, why doesn't the weight of all the water above keep it down?

ANSWER: As you go deeper and deeper the pressure in the water gets larger and larger. Suppose that you have a cube of wood. So, the force down on the top of the wood is enormous
so you might think this would certainly hold it down. However, if you go a little bit deeper to the bottom of the cube you find the pressure is a little bit bigger than at the top, so the force up on the wood is a little bit bigger
than the force down at the top.

QUESTION:
Witch falls faster a bowling ball or a tennis ball

ANSWER: In a vacuum, both fall the same. In air the force
F due to air drag, which points upward if the
ball is falling, may be approximated as ¼Av ^{2} where
A is the area presented to the onrushing air and
v
is the speed. The net force on a falling ball is mg (the weight of the
ball) down plus the drag force, ¼Av ^{2} -mg=ma ^{
} so the acceleration is a =[¼Av ^{2} /m ]-g .
Putting in the size and mass of the two balls I find
that the drag accelerations are a _{tennis} =0.015v ^{2}
and a _{bowling} =0.0013v ^{2} .
The drag causes more than 10 times the acceleration on
the tennis ball which will therefore fall more slowly
than the bowling ball.

QUESTION:
Momentum is always conserved, whenever I hear a
statement that momentum is not conserved, it is usually
because the system being evaluated is not big enough to
account for all the momentum. With that said, I often
hear statements that momentum is not conserved in
General Relativity, but if you account for the momentum
in the field, momentum is conserved. maybe not in the
mechanical mass x velocity aspect, but if you account
for all the momentum, including field momentum it
appears to all be accounted for. So, with all that said,
why do you read / hear statements about how General
Relativity does not conserve momentum? Is it because
they are only viewing the mechanical mass x velocity
momentum? And not taking into accounting for the
momentum in the gravitational field?

ANSWER: Details regarding
general relativity are beyond the scope of this site; I often answer broad questions on that subject, the general principles (principle of relativity, spacetime, equivalence principle, etc .) Let me tell you
what I do know about linear momentum and then I will point you to a link where you can learn more about your question.

In Newtonian
mechanics, a vector quantity equal to the rest mass
m times the velocity v is
found to be constant for isolated systems ; d(mv )/dt= dp /dt =F _{ext} .
The quantity F _{ext}
is the net sum of all forces acting on the system from
the outside; similarly, p is
the vector sum of all linear momenta of all the pieces
of the system. This equation is just a different way of
writing Newton's second law, so you may think of linear
momentum conservation as a clever trick, not some new
physical law.

In classical
electricity and magnetism, you are right: if you want to
see familiar conservation princlples you must consider
the energy, linear momentum, and angular momentum content
of the fields.

In special
relativity, if you define linear momentum as rest mass
times the velocity you find that it is not conserved for
isolated systems. It turns out that you have to redefine
linear momentum as γmv it is conserved
where γ =1/√(1-(v ^{2} /c ^{2} )).
I have discussed this here many times before and if you
go to the faq page and search for momentum you will find
several links. A more sophysticated way to view all this
is in terms of 4-vectors where the energy-momentum
vector has the (new) linear moment as the space-like
components and energy as the time-like component.

Finally your
question, what about general relativity? From my brief
reading it seems to be a lot like electromagnetism—when a particle is bent by gravity, thereby changing its momentum, the field is also altered. General relativity is a very mathematical
theory and so any discussion of details is likely to be
beyond the purposes of this site. There is a good
readable discussion at this
link .

QUESTION:
I am sure you know that if you drop a hammer and a feather in a vacuum from an equal height, they will both hit the ground/floor at the same time.
This I belive is because gravity exerts the same force on both objects but the feather has more air resistance to weight ratio.
I drive trucks for a living so forgive me if this sounds dumb as I have no scientific training.
My question is that once they are on the ground, why is the hammer harder to lift than the feather?
I understand the hammer weighs more, but isn't weight just a consequence of the pull of gravity?
The hammer weighed more when it was falling but was attracted to the earth's centre at the same rate as the feather (in a vacuum)

ANSWER: Gravity does not exert the same force on each, it exerts a force (called its weight) which is proportional to
the mass of each, w=mg for the feather
and
W=Mg on the hammer where g i s the acceleration due to gravity.
Now, Newton's second law tells us that if an object with
a mass m (or M ) experiences a force
w
(or W ) it will experience an acceleration a=w /m=mg /m=g.
(or A=W/M=Mg/M=g ). So you see, with
no other forces they will have the same acceleration
which means they would speed up together thence hitting
the ground simultaneously. In words, mass is a measure
of the resistance to being accelerated by some force and
weight is a force which is proportional to mass, then
all masses fall with the same acceleration. In the real
world there is air, and there is therefore air drag
which is approximately proportional the the square of
the speed and points upward, but it does not depend on
the mass. Therefore the larger mass is less affected by
the drag and wins the race as you know it must!

QUESTION:
Since all objects that have mass -the earth, a book etc. - cause a distortion of spacetime resulting in gravity, and E=mc2 tells us that mass and energy are the same thing, shouldn't energy cause a distortion of spacetime resulting in gravity?

ANSWER: You are absolutely right! Spacetime is distorted by energy density.

QUESTION:
I'm a forensic scientist and feeling a bit out of my
league here. I'm working on research to combat specific
junk science ("grave dowsing"). I've published a blind
study that showed the technique ineffective but one of
the proponents continues to espouse what are clearly
erroneous statements about a piezoelectric property of
bone and oscillation of electromagnetic fields that he
claims cause the divining rods to move. I've only seen
one thing published about this, a really horrible
article published out of the University of Lulea,
Switzerland that had some overt and unforgivable errors
in design. I think just a general understanding of
simple physics make it obvious that even if bone had
piezoelectric properties (which I very much doubt), that
the electromagnetic field produced would be able to
affect metal rods "1/2 mile away" as this instructor is
telling students. I could do a bunch of experiments to
disprove this but I think the answer is already there in
physics and associated mathematics but I don't possess
anything near the understanding of the mathematics to
work the equations. My question, if you can forgive all
the buildup, is whether you think I'm on the right track
looking at the Lorentz force and Maxwell's equations.
Would I be able to use those equations to figure the
amount of charge that would have to be produced by a
bone in order for it to move a metal rod even a meter
away?

ANSWER:
Dowsing is a really tricky thing to try
to analyze. There are some very bright people who thinks
that dowsing for water works. And of course water is not
going to exert a force on a willow branch or coat
hanger. My feeling is that, like the old ouiga board,
that the pushing of the stick is done by the dowser. And
it is altogether likely that he does this unconsciously;
these are often old codgers who have been doing this
most of their lives and have experience. Another
suggestion is that the some of the successes of water
dowsers are that they are often working in areas where a
drilled well anywhere in the vicinity would probably
find water. So, now, what about the grave dowsers? I
have found that bones do indeed have piezoelectric
properties; it plays important roles in bone formation.
It is not clear to me if it has to be a living bone, but
just let us suppose that the weight of the ground above
pushing down on the bone (a person in a coffin would not
have any pressure being applied to his bones) creates a
potential difference. The thing is that the net charge
created is zero, the bone becomes positively charged on
one side and negatively charged on the other; in that
case, it looks like a parallel plate capacitor and
nearly all the electric field would be from positive to
negative, nearly all the field being confined to the
bone. And, because it is dipolar, whatever field exists
outside the bone would fall off very quickly (faster
than the field of net charge which falls off like 1/r ^{2}
with distance r ); the field the dowsers think
they can detect would fall off more like 1/r ^{3} .
Add this to the fact that piezoelectric fields are very
weak in the first place, I can see no possibility that
they could be detected even by the most sensitive of
instruments, let alone a piece of a coat hanger.
Devotees are very serious about this, though, and it is
a losing battle to try to convince them otherwise.
Collect as much data regarding success/failure rates, do
a good statistical analysis, and be happy with that.

Incidentally,
statistical analysis is vital. The classic example is
from several decades ago. For years it was presented as
fact that clusters of cancer occur in communities where
high voltage power lines crossed over, presumably due to
the electric and magnetic fields associated with them.
This was totally debunked by doing proper statistical
analyses.

QUESTION:
If the angular and the linear velocity of a wheel at the point of contact to the road is equal to zero.
Where does the velocity or the kinetic energy of the still moving car all go?

ANSWER:
It didn't "go" anywhere. There is no rule that anything which has kinetic energy must have all parts of itself moving with the same velocity.
Those parts of the tire in contact
with the road contribute nothing to the kinetic energy of the whole car.
Perhaps it is a bit easier to understand with a WWII
tank. The tread on the ground is at rest; the tread on
top has to be going faster than the tank as a whole; all
the wheels and gears are spinning and different parts
are going up or down or forward or backward or in all
other directions with different speeds. But, if you add
up every single part's contribution to the kinetic
energy of the whole tank, it will stay the same if the
engine provides enough energy to overcome all the
frictional forces present.

QUESTION:
Suppose you have two massless containers that are empty. They have no mass so there is no charge nor gravitational attraction between the containers.
If we keep adding protons and neutrons into each container equally, a coulomb attraction from charge and a gravitational attraction due to gravity will develop. Initially, the coulomb force will far outweigh the gravitational force. However, as we continue to add protons and neutrons to the containers, the amount of mass collected will grow to the point where the macro size is large enough to no longer have much coulomb attraction.
However, the gravitational attraction will grow.
Is there a point where the attraction of charge will equal the attraction of gravity. Has this been investigated? And has this been achieved?

ANSWER:
I have held back thinking about how to answer this question for some time now. Only this morning did I realize that it is impossible. The reason is that if each
container has a mass M and a charge Q , what determines the relative magnitude of the forces between the two containers due to gravity and to Coulomb
forces is only going to depend
on the ratio M /Q ; but you propose to
add equal amounts of charge and mass to each container
so this ratio would never change!

There is also the
problem of the interaction among the nucleons in the
containers. You are essentially imagining creating super
heavy nuclei and in any nucleus the mass is not equal to
the sum of its parts. It would be impossible to
determine the mass of systems of neutrons and protons in
the numbers far above natuarally occuring nuclei

What I can do is
calculate the ratio for two hypothectical point masses
(M ) and point charges (Q ) separated by a
distance R such that the attractive
gravitational and repulsive Coulomb forces have equal
magnitudes.

GM ^{2} /R ^{2} =kQ ^{2} /R ^{2}

M /Q =√(k /G )=√(9x10^{9} /6.67x10^{-11} )=1.16x10^{10}
kg/C.

QUESTION:
Let's say there is a container that contains water and an ice cube in floating in it. Some portion of ice is below the water (may be referred as bottom of ice) and some above (may be referred as top of ice). Now volume of only submerged portion of the ice is equal to the volume of the water displaced, isn't it? If so, where did the volume of melted upper portion of the ice that did not displace the water previously go? Should it not be added, hence increasing the total content of water? But that's not the case according to text books.

ANSWER:
The reason is that the ice does not have the same density (less than) as the water; that's why if floats. But, if it melts it become water and therefore has a density the same as the water.
The fractional amount by which the density decreases is precisely equal to the amount by
which the volume decreases.

QUESTION:
My question has to do with wind chill. I believe that I understand the basic mechanisms by which this works, but my question gets more convoluted. Thinking about how objects which fall into the earth's atmosphere from space will burn up due to the extreme heat of friction with the air, I am wondering if it is possible to say at what velocity would wind change from having a chilling effect to having a heating effect due to friction? My guess is that it would require wind speeds which would far surpass the point of being fatal to a human being, thus negating any specific concerns about wind chill as such,
but who knows?

ANSWER:
First of all, you are extrapolating a
qualitative quantity (essentially "how cold do you feel
in a wind") into a region where it was never meant to be
applied. You can find lots of details about wind chill
in the
Wikepedia article , but this article also states "Windchill temperature is defined only for temperatures at or below 10 °C (50 °F) and wind speeds above 4.8 kilometres per hour (3.0 mph)".
The reason is that this effect of wind is to hasten
temperture loss of the body due to the wind, and is most
certainly not applicable to situations where the effect
of wind is to increase, not decrease, temperature. In
the summer time the main reason we feel hotter or cooler
is the humidity. Instances where air drag has a
nonnegligible effect of heating or cooling, these ideas
do not apply. There is, in fact, no way to have a simple
qualitative formula for the effect of speed on heating
because it depends on the shape and size of the object
and on the density of the air through which it is
moving; the SST, for example had a "needle-shaped nose"
to minimize heating at supersonic speeds and would have
burnt up otherwise. The takeaway here is that
qualitative quantities like this may play an important
role in health and safety (e.g ., high winds in
cold weather can result in high probability of
frostbite) and provide information helpful to deciding
how to dress, for example, in different wind conditions.

QUESTION:
Hi. My question has to do with gravity. If you took two planets with the same mass, in an area of space without any other immediate gravitational influences, and put them just within each other's gravitation pull but not moving, would they collide, or would they orbit each other? If they would end up orbiting each other, why?

ANSWER:
As long as these planets are spherically symmetric (center of mass at the center, density only dependent on distance from the center), their centers will accelerate with equal accelerations
until they collide. They could either stick together and come to rest or bounce off each other and each recoil with equal speeds in opposite directions.
If they were not spherically symmetric, their centers of
mass would certainly not be exterior the the planets, so
their centers of mass would accelerate towards each
other and they would still collide and therefore not
orbit; if they stuck together, they would rotate about
their total center of mass but have no translational
kinetic energy.

QUESTION:
If two objects move parallel at the same speed light years away would the a person on one object be able to see the other object? This is assuming a similar starting position, my initial assumption would be no as any light from one object would fail to reach the position of the second object. Or would the two objects be able to observe each other but one would appear "behind" the first? I know its multiple questions but what would be the shift of the object if the second question is in the positive?

ANSWER:
Yes, the light from the trailing object would reach the leading object Y years after it was sent where Y is the number of light years between the two.
You are not thinking about this problem using the
principles of special relativity (SR). Perhaps the most
important assumption of SR is the principle of
relativity which asserts that the laws of physics are
the same in all inertial frames of reference. An
inertial frame is one which moves with constant velocity
relative to another frame in which you know the laws of
physics. In your case, since both objects are moving
with constant speed in the same direction, they might as
well be at rest.

QUESTION:
Is anyone trying to figure out how to tap into the energy that holds atoms together? I don't mean harnessing it to fuel our energy needs. I mean accessing it like we would a hard drive.
It's the one thing that everything in the universe has in common and it's the one thing that connects us all. In it is knowledge beyond comprehension.

ANSWER:
So, the question you are asking is very vague. For example, what does "holds atoms together" mean?

Does it mean what holds atoms together in a macroscopic piece of matter?

Does it mean
what holds the atom itself together?

Does it mean what holds the nucleus of an atom together?

And what does "accessing" mean?
What we often think of when something is being held
together is a force; in more advanced physics we usually
express the presence of forces by introducing potential
energy functions. So I am going to answer your question
with a very broad brush; in essence, to "access"
anything at the atomic level you need to ascertain, by
doing experiments and interpreting the data, what the
relevant potential energy functions are (or forces, if
you like). And, guess what! That is what physicists and
chemists have been doing for the past several hundred
years. Condensed matter physicists address #1 above.
Atomic and molecular physicists address #2. Nuclear and
particle physicists address #3. In fact, thinkers from
thousands of years ago, Greek philosophers for example,
have been puzzling over these kinds of questions. So, is
"anyone
trying to figure out" answers to these questions? You
bet! Thousands of men and women for thousands of years!

QUESTION:
Can air pressure be different inside the same container? I'm having a debate with my work. WE have a device (packer)that inflates like a long cylindrical ballon inside sewer pipes to install fiberglass patches. They are claiming that if our packer is at the end of the pipe and sticks out at all, that the air pressure inside the pipe, and contained, is less than the air pressure that is outside of the pipe and is bulging out because it is not contained (within the pipe). I say that the pressure is constant throughout the entire thing and there is no change in pressure inside vs outside the pipe because pressure has to be constant inside one container. Who is correct?

ANSWER:
So, lets assume that there is some pressure
P inside the balloon and it has some volume
V and that it is wholly inside the pipe. Now it
comes to the end of the pipe and starts expanding
outward since it no longer has the pipe to keep it to
its previous cylindrical shape. If this happens pretty
quickly the temperature will approximately stay the
same. So the product of P and V will
remain constant as it expands. The volume will increase
so the pressure will decrease. During the time this
expansion is happening, there will be a different
pressure inside the pipe and outside but that will be
for a very short time, the time it takes for the
slightly lower pressure to equilibrate throughout the
volume. If everything is static and in thermal
equilibrium, the pressure will be the same everywhere.

QUESTION:
I have a question pertaining to sound. So, I was
wondering how it is possible for me to hear someone
facing away from me in say, an open field. Because I
know for example, if in a closed space, the sound would
bounce off the wall and come back to me, at least that's
how I think it works. But if I'm able to hear someone
facing away from me in an open field, Is sound more
complicated than a directional wave, is it more spread
out than I think?

ANSWER:
Usually in physics we think of wave
sources in some simple geometric form. For example, if
you think of sound as just like a laser for light as a
narrow coherent beam of sound coming out of your mouth,
only someone standing in front of you would be able to
hear you, clearly not the case. If you think of it as a
point source but radiating only forward, anybody in
within a hemisphere in front of you would be able to
hear you. But, as you note, this second simple model
does not fit evidence from the real world, so it must
also be incorrect. But if you think of it as a simple
point source it would radiate in all directions. But
that would say that I hear you just as loud whether in
front of you or behind you; this also is not what we
experience which is your voice is less loud if I am
behind you rather than in front of you. However, we
could account for the smaller amplitude from be due to
the sound source being much more forward in the
head/throat so sound emitted backwards gets attenuated
passing through the head tissues and bones. The fact is
that the source of a human voice is complicated; there
is the larynx containing the vocal cords, the primary
source of sound; but then there are also resonant
cavities*, the mouth, nasal structures, sinuses, etc .,
which amplify and rebroadcast the sound.

*If you just had a
guitar string simply stretched between two points in a
room, you would be hardly able to hear it. The "box" of
the guitar plays a vital role by resonating and
amplifying the sound from the vibrating string.

QUESTION:
I've been thinking about the rising prevalence of coilguns, and an interesting puzzle came to my attention, which it is unfortunately beyond my limited "physics" ability to solve.
Some context:
Rifling in conventional kinetic firearms use grooves in the barrel to change some forward momentum into rotational energy, affording stability to the projectile.
However, I want to try to achieve, with magnetism, an inverse effect - converting rotational energy (at least partially) into forward momentum, without any physical contact with the rotating entity in question.
Assuming no constraints imposed by current technological progress:
What forces, where, and when would be required to achieve such an effect?
Feel free to go into any level of technical detail that you feel is necessary.

ANSWER:
You did not mention why this is of
interest to you, so let's assume, since you mention
guns, for purposes of an example, that you would like to
take a rotating bullet-like object and convert
its rotational kinetic energy into translational kinetic
energy. Consider a solid cylinder with mass m , angular velocity
ω , radius R , moment of inertia I =½m R ^{2} .
So, it will have a rotational energy K =½Iω ^{2} which we wish to convert 100% into translational kinetic energy
K =½mv ^{2} . If you do
all the requisite algebra you will find that ω =(√2)(v /R ).
Suppose R= 1 cm=10^{-2} m, a very big
bullet, and v =1000 m/s, a modest bullet speed;
then I find that ω =1.4x10^{6} rpm.
More than a million rpm means that this would not be a
good way to get energy to launch a projectile.

QUESTION:
Baseball commentators often say a hard throwing pitcher is "providing the power". They are meaning the batter doesn't need to swing as hard as he would have to if the pitch was slower. I don't think that can be correct since the faster pitch will have more momentum in opposite direction of the force provided by the bat. What is the correct way of thinking about that?

ANSWER:
I believe that the best way to think about this is using impulse. The impulse is the average force
F times the time Δt which the ball and bat are in contact,
I=F Δt .
Because of Newton's second law, F =Δp /Δt , I =Δp where Δp is the change of linear momentum. If the incoming pitch has a speed of
v _{1} and the outgoing ball has
speed v _{2} , the change in momentum is
m (v _{2} +v _{1} )
where m is the mass of the ball, and therefore
v _{2} =(I /m )-v _{1} .
So you can see that the faster the pitch, the slower the
speed of the hit ball if the batter exerts the same
impulse to the ball. I suspect that the reason the commentators
believe this is that a fast ball right down the middle,
often called a mistake on the pitcher's part, probably
yields the most home runs since it is the easiest for
power hitters to hit; the reason is not that the ball is
going faster, it is because it is going straighter. (All
this assumes that the outgoing and incoming velocities
are in exactly opposite directions to make my example a
one-dimensional problem.)

QUESTION:
I want to demonstrate that NaCl solutions conduct electricity to my students and show this qualitatively and quantitatively.
Is it possible to build a complete circuit as follows? |--9 volt battery---NaCl solution---multimeter---LED/Daylight
bulb---|
|---------------------------------------------<-----------------------------|

ANSWER:
Assuming that your multimeter can measure current as well as voltage
this should work; if you are going to measure
conductivity you need the current as a function of the
concentration. Also, I would do the quantitative and qualitative experiments separately because LEDs are not ohmic
and as you change the current you would change the
voltage across the LED and therefore across the
solution; you want to keep the voltage across the
solution constant at 9 V when you are measuring the
conductivity. Also, the resistance of the filament of an
incandescent bulb changes as it heats up, so you should
only use that when you are doing the the qualitative
demonstration. Come to think of it, an incandescent bulb
would probably be better because not all LEDs are
dimmable so it might just turn off when you changed the
solution or not change at all.

QUESTION:
Jupiter's moon Io is volcanically active due to the moon being squeezed by the gravity of Jupiter and the interactions with the other moons. Io is hot, where is that energy coming from? There is not supposed to be a 'free lunch' regarding energy, there has to be a cost in energy somewhere else. So what is the supplying the energy in Io. Gravity isn't an energy source, so what is the source of energy for Io?

ANSWER:
I usually do not answer questions about
astronomy/astrophysics/cosmology, but will give this one a
try. Why do you say that gravity "…isn't an energy source…"? It is one of the most important sources of energy there is; a ball dropped increases its kinetic energy as it falls
because of the force of gravity doing work on it. In the case of Io, there is a large
tidal force
because of its gravitational interactions with Jupiter
and other moons which are constantly squeezing and
changing its shape. Imagine that you are squeezing and
relaxing a rubber ball; the energy you are putting in
will cause it to heat up. For a full-blown explanation,
see the Wikepedia article on
tidal heating .

QUESTION:
Why does certain object falls faster than others?

ANSWER:
If there is no air, all objects fall with
the same acceleration. However, if the objects are
falling in air, the air drag force affects some more
than others, and they fall with different accelerations.

QUESTION:
Why light speed is a universal constant if gravity can bend space-time?

ANSWER:
In everyday life speed and velocity are
synonymous. In physics velocity is a vector (specified
by a magnitude and a
direction) whereas speed is the magnitude of the
velocity, a scalar. The speed of light is a universal
constant, the velocity need not be. So a beam of light
may be bent by gravity but still maintain a constant
speed. It is like driving on a curvey road at a constant
speed of 50 mph: your speed is a constant but your
velocity is not.

QUESTION:
I have a weird question I'd like to ask. I was in the bomber ride, which consists of a long arm with a gondola seating 4 people at each end. The arm revolves forwards for a few times then backwards for a few more times, and the gondola also rotates 360° mid air. The arm is 120 feet high and revolves at a speed of 13 rpm, in which the passenger experiences 3.5 Gs according to Billy Danter's fun fair website. I vomited a little while the gondola was at the very top and rotating, but I cannot remember if we were going forwards or backwards. I apologize for the gross details, but I'm curious to know where my vomit had landed, as it did not splatter on me not on the next seat passenger.

ANSWER:
T he
picture of the ride is shown on the left. It rotates (13
rpm) about an axis in the center. If you want to see a
video of this, click
here . If you are at one of the outer ends of this
thing, you will experience a "fictional" centrifugal
force which is 3.5 times larger than your weight (I
checked and this is correct). If you watch the video you
will see that the gondola itself is also free to rotate
about its own axis, but any centrifugal force due to
that rotation is trivially small compared to the main
centrifugal force and your own weight. The other picture
shows the gondola at one end rotated to some angle and
at the very top as specified by the questioner. The
questioner vomits (I will assume straight out) and the
vomit is labelled V and has a velocity
in the direction of the yellow arrow. There are two
forces, the orange vectors, on the vomit, its weight
W down and the centrifugal
force 3.5W up. The net force
on the vomit is therefore about 2.5 times its weight and
so it will appear to someone in the gondola someone in
the gondola to follow a path roughly like the
green-dashed line in the figure. So it is no surprise
that it did not get on anybody.

QUESTION:
I have recently gotten into synthesizer and how they affect soundwaves. Can soundwaves really be a Sawtooth and triangle's waves? The sine wave is the only wave that seems possible to me.

ANSWER:
I don't know on what you base your belief that only sine
waves are possible. In virtually all musical instruments
the sounds produced are not pure sine waves; only
electronic instruments are capable of creating simple
sine waves. Imagine two instruments, say a violin and a
piano, both play a middle C; do they sound just the same
to you? Of course not; this qualitative difference is
called timbre . The pitch of a tone is
determined by its periodicity, the frequency with which
the wave repeats its shape when passing; not only sine
waves but any shape you can imagine could repeat itself
over and over. Your brain hears the middle C as the same
note from all instruments but hears differences as well.
But, here is where you have gotten something right when
you think of sine waves as being somehow special.
Physicists and mathemeticians find sinesoidal waves to
be very easy to work because they satisfy a very simple
differential equation and because they form what is
called a complete set of functions. As a result, any
periodic function with frequency f may be described
perfectly by a sum of sine and cosine functions with frequencies
f ,
2f , 3f , 4f …
This is called a Fourier series representation
of the wave. The sum is infinite, but normally only a
few members are needed to give a good representation of
the wave. The figure shows Fourier representations of a
square wave for 1, 2, 3, and 4 terms in the series.

QUESTION:
What causes one person to float while another person only sinks? Is it the molecular make up of the water, the human, or a combination of both? I am very buoyant, I lay on the surface of water, even with a weight belt is hard for me to stay under water. My GrandMother floats like a large rock, straight to the bottom, she can't even swim with a life vest on. Why is that?

ANSWER:
The answer is pretty simple: if the density of the body
(mass/volume) is less than the density of water (997
kg/m^{3} ) it will float, if more it will sink.
The body is made up mainly of bone (~1900 kg/m^{3} ),
fat (~914 kg/m^{3} ), and almost everything
else (~1040 kg/m^{3} ).
Approximately 14% of body mass is bone. So, only fat has
a density less than water meaning the more fat you
carry, the more buoyant you are likely to be. It is
difficult to try to get more quantitative because the
lungs occupy a not negligible fraction of the body
volume and air density is ~1 kg/m^{3} . Almost
nobody with air in his/her lungs will sink like a stone.
Your grandmother probably has very low body fat.

QUESTION:
Thank you for taking the time to read my question. I am seeking an
explanation to my problem from some time now and I am very glad to have
found your site. I'm an olympic wrestling strength and conditioning couch
and I have my athletes practice an old isometric exercise called pushing the
wall, to develop whole body strength and power. As I have gone thorough this
exercise myself, after a year or so I have discovered that something
interesting is happening and this is what I'm seeking an explanation for.
As I push on the wall ( the push really comes from my back leg and the whole
body but without whole body muscular tension), with the same pressure,
without any other movement ( or an immovable surface) for let's say 30sec -1
min' when I slightly "push off" the wall ( I'm about 96 kg) and I mean
gently, not with force, I find myself been thrown backward for 10-15 meters
or more in some instances, by some kind of force..For the first few meters
the force is very strong that I can't stop the movement myself. The momentum
actually accelerates and it gets stronger as I keep been pushed backwards
and my body keeps going backwards, after a while the momentum comes to a
stop by itself, if I don't make an effort to stop it myself. Even if I put a
slight pressure on the wall and push off very gently, the same effect takes
place.

REPLY:
I have spent some trying to understanding how this could be. I keep coming
back to your question but to no avail. If you still want me to try to figure
this out, I would appreciate it if you would send me a video you mentioned.

FOLLOWUP #1:
Thank you so much for getting back to me. I really thought that my inquiry was weird enough for you to not consider it as true. For a while now I considered asking or talking with a physics professor about this and see if there is an explanation for this. You are the only person I thought to ask, and because I think this is a strange phenomena, I thought you might think it is fake and not bother taking it seriously. Because I think is indeed strange ( I call this force: "strange power"), I only think what other people might think about it and not even bother with it, as it seems that it is not a tangible physical thing or force, meaning that it seems that it isn't based on muscular force or physical exertion. As you see in the videos I really don't push hard at all.

ANSWER:
Let's look carefully at one of
the videos
you sent. I have inserted one frame showing you right
after you have lost contact with the post; your body is clearly tending
toward a sitting position. Your weight (one of the
forces on you) acts at your center of gravity, clearly
behind your feet. If you don't do something, you will
fall on your ass because your weight exerts an
unbalanced torque about your feet. To keep that from
happening, you push forward with your feet on the floor
and the floor pushes backwards on you; if the floor were
extremely slippery, e.g . like ice, there would
be nothing you could do to avoid going down. But, there
being friction, you now have an unbalanced force on you
which accelerates you backward. This, in itself, does
not keep you from falling but it gives you time to
straighten back up so that your weight no longer exerts
any torque about your feet and you can now stop without
fallin g. There is no "strange power" force
here, just plain old Newtonian physics.

FOLLOWUP #2:
Thank you for examining that frame from the video. But, what is that, that is strongly moving my weight and acting on my center of gravity ( as you see in that frame) ?
There is a strong power/force that is causing my position that you see in that frame, my center and body weight gets pushed backwards by that force and it keeps pushing me backwards while I step backwards trying to regain balance and stability. But, there is a strong force that strongly pushes me in that position, and keeps acting on me, when I only slightly push on the wall ( as you see in the videos). What is that force? and where is it coming from?..that is my question. What is that initial force or power that throws me backwards ( which is not the momentum caused by my center of gravity being behind my heals, there is a power that first pushes my weight and/or center of gravity in different positions thus creating a movement afterwards ). Take a look at my starting point. There are almost no external moving parts, that will logically/physically initiate a strong movement of my weight or center of gravity backwards and yet it happens.
And also how can the other things that you see in the other videos be explained.

ANSWER:
I will answer your last question first. You sent four
videos, each with 3 or 4 trials. In every trial, just
before you lost contact with your hand, you swung your
hips back so that your legs were angled backward so that
your center of gravity (somewhere inside your chest) was behind your feet. There
are no "…other
things that you see in the other videos…" which
need to be explained beyond this single frame and the
little backward scurrying that follows.

I
thought that my answer to your first followup question
pretty clearly explained the whole thing without having
to get into the details of the forces and torques I was
talking about. But, it appears from your second question
that you are not familiar with how forces are dealt with
in physics. So I will give a little tutorial on the
important points necessary to understand your situation.
The most important thing to understand are Newton's
three laws:

If an object
is at rest or moving in a straight line with
constant speed, the sum of all forces on it
must add to zero.

If an object
with mass M has net force F not
equal to zero it will have an acceleration a=F /M
in the direction of the net force.

If an object A
experiences a force F from another object
B, object B experiences a force of equal magnitude
but in the opposite direction as F .

As an example, I will look at a case very similar to
yours, a sprinter. The figure shows all the forces
acting him, his weight W which is the force
which the earth exerts vertically down on him, the
normal force N which the ground exerts up on him, and
the frictional force f which the ground exerts
forward on him. Since he is not accelerating in the
vertical direction, only forward, N must be
equal in magnitude to W so that their sum is
zero. The net force at this instant is f , a
forward force which accelerates him in that direction.
Most people will say that it is the muscles in his leg
which propel him forward, but that is wrong; the muscles
in his leg result in the foot pushing backward on the
floor so, because of the third law, the floor pushes
forward on him. (Most people would say that an engine is
what drives a car forward, but it is actually friction
between the tires and ground which causes the car to
accelerate forward. You might be interested in an
earlier
answer .) There is still a subtlety regarding this
situation which is important: suppose this sprinter
encounters a patch of ice so that the friction
becomes
essentially zero. Now there is a net force on him which
is zero. Does he simply glide across the ice and then
resume running when he gets back to dry land? No,
because his weight will cause his whole body to rotate
about an axis passing through his feet and he will fall
forward until his other foot or his knee hits the
ground. (We say that W exerts a torque equal to W
times the distance horizontally between the axis and the
center of mass.)

So now your case. The next figure shows you a short time
after you lost contact with the pole. You had to do
something or else you would rotate because of the torque
due to W and end up sitting on the ground*.
Your brain can detect this situation and lifts
your right leg and pushes on the floor with your left
foot, causing a frictional force ("strange power"?!)
pushing you backward and accelerating
you.
You will still be starting to rotate but note that your
right foot will hit the floor less separated from the
weight so your tendency to fall will be lessened due to
the decreased torque. When that right food lands on the
floor the left foot lifts and the roles of the two feet
reverses. You keep backpedaling like this until,
finally, your center of gravity is directly above your
feet and you won't fall over.

*In one of your trials you failed to be able to stop the
rotation and did, indeed, land on your ass. Notice your
arms and lifted leg desperately trying to move your
center of gravity farther relative to your right foot
but it did not work! The reason was apparently that you
dropped your hips too far when you launched.

If you just stood straight up with your center of
gravity vertically above your feet, you would not feel
the "strange power" when you lost contact with the pole. I know—I
tried it.

FOLLOWUP #3:
I do understand what you are saying. Your last sentence was : "If you just stood straight up with your center of gravity vertically above your feet, you would not feel the "strange power" when you lost contact with the pole. I know—I tried it." Well, I tried it too, and again the "strange power" was pushing me off the wall..( strange power...is only a joke :)
The upper body gets pushed by the force first and the feet follow. It looks like the feet are trying to catch up with the momentum of the upper body, but it doesn't feel that way because the force keeps pushing the upper body backwards not because of the momentum of the upper body going backwards. Again, it is the force, a very tangible force, from the slight push off the wall that keeps sending the body backwards and that creates the acceleration. The acceleration of movement doesn't come from trying to stay on my feet or to regain stability after the push off the wall. I know it looks like that but the feeling is not at all like that. This is how it is felt. Also sometimes the force goes down into the feet, making them move backwards, thus moving the whole body backwards, Anyway, maybe it is difficult to understand. It is something that has to be felt, and I have tried to explain it but maybe it is not clear in how I say it. Maybe the only explanation using physics is the one you are giving me. ( I was thinking more in terms of conservation of mechanical energy, potential energy stored in the mechanical bonds between atoms, stored elastic potential energy that changes into kinetic energy when it is released )
Here is a video of it, me standing upright. If you are insisting on the explanation of Newtonian laws at work here, I understand and I thank you for your time.

ANSWER:
Here are the first few seconds of one push. Your feet stay put but your body leans. By the time of the fifth picture your body feels like it is falling and your brain perceives you are falling. Your brain interprets that something is pushing you and screams to your body that it has to act fast, keep one foot planted and get the other foot back to try to stop the fall. And so forth as described in my answer.

FINAL
FOLLOWUP: Yeah, ok. I was sure you would say that. Your opinion makes sense when only watching, yet you seem to reject what I am trying to explain to you,
that what you see is not what is actually happening, ..but you can have your opinion and thank you for that.
It is great that you gave me your point of view, because now I know what other people might say and I know what to expect.
If you would feel what I am talking about you would not have the same opinion. But it is perfectly ok.
Thank you for your answer, thanks for trying.

MY
COMMENT: This is certainly not the first time I have had to end with "we'll just have to agree to disagree)!

QUESTION:
I built a spring loaded catapult which sent a practice golf ball a distance of 11.25 feet from a height of 2.66 ft at the horizontal (0 deggres) which I believe gives a velocity of 27.64 ft per sec, since free fall time is 0.407 sec. Then I launched the ball from a 28 deg angle and my assumption is that the initial velocity from any angle would now be 27.64 ft/sec, however the distance only came to 19.25 ft when I predicted it should go 23.8 ft. Working from the distance of 19.25 ft I get an initial velocity of 24.27 ft/sec. How come the initial velocity at another angle is different than the velocity when fired horizontally? I am doing this project in my high school Pre-Cal course of which I am the instructor.

ANSWER:
This will be a good lesson for your
students. You are assuming that the usual kinematic
equations for motion in a uniform gravitational field
are exactly correct. But they are only approximations.
In the real world there is air and air drag causes a
force opposite the velocity vector which is
approximately proportional to the square of the speed.
So the mathematics are going to well beyond high school.
Let's just consider the first case to discuss here since
I can make rough calculations to estimate the effect of
drag. I will not go into details, just give you the
results of my scribbling. First you got a velocity 27.64
ft/s by dividing 11.25/0.407; this is only the
horizontal component of the velocity because the ball
would also acquire a downward vertical component of
0.407x32 ft/s=13.02 ft/s. The velocity when it hits the
ground is √(13.02^{2} +27.64^{2} )=30.55
ft/s. But that is if you ignore air drag which will take
velocity away as it flies. My rough calculations are
that if the ball has a speed of about 30 ft/s the drag
will exert a force of about 0.013 lb. This will
cause an acceleration of about -4.3 ft/s^{2}
which would result in a loss of velocity over 0.407 s of
about -1.75 ft/s. (What makes the math difficult is
that, unlike uniform acceleration by gravity, the force
on the ball changes with time.)

The lesson your
students should learn is that it is important to ask
yourself how good a description any "theoretical"
model of a physical situation like projectile motion is.
Your results were pretty good, but you were dealing with
speeds of only about 20 mph; a driven golf ball can have
speeds up to about 200 mph so the drag force they
experience would like 100 times larger than at your
speeds (proportional to v ^{2} ).
Predictions of the distance of a drive would not be close to how far the ball would actually go.
Another example is a baseball; anyone having played
outfield knows that a long fly ball ends falling almost
straight down, not at the same angle it was launched.
You can demonstrate this by throwing as hard as you can
a crumpled sheet of paper; it will end up falling almost
straight down having lost all its horizontal velocity.