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Search Advanced search…. Log in. Support PF! Buy your school textbooks, materials and every day products via PF Here! Homework Help Introductory Physics Homework. JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding. Calculate momentum of the puck. Thread starter rrosa Start date Apr 24, The hockey stick is in contact with the puck for 0.

The puck is not moving before the player hits it. The attempt at a solution I got 0. Shouldn't the answer be the same for both. I really don't understand the difference between both questions. I don't know the math, but is there a certain height that this platform could be built so that the jump is possible? How high can you jump from on earth, maybe 7 m about 20 ft?

Certainly not "hundreds of feet", more like ft.

Rotational inertia (article) | Khan Academy

They sure look that way on first inspection and the reason they are taught to be two different things is because of history: it took a long time to find the connections between them. Two centuries of experiments finally led to Maxwell's equations , four equations which show all the ways the electric and magnetic fields are related. Maxwell's equations comprise a relativistically correct theory of electromagnetism; it was actually questions like you ask which led Einstein to discover the theory of special relativity. In its most sophisticated form, we think not about electric and magnetic fields but the electromagnetic field.

The electromagnetic field is not a vector field, rather a tensor field, and any field can be decomposed into its electric and magnetic parts. Your friend on the ground will also see an electric field but it will be slightly different from the field you see, which you may want to interpret as some of the electric field transformed into magnetic field. I found an estimate of the current amount of nuclear waste, about 70, tons.

The shuttle had a weight of about tons and our biggest rockets could put it in a near-earth orbit, not totally escaping earth's gravity. So, it would take something like 50 launches just to get rid of what we have. The second, and perhaps more important, reason is that it could be an environmental disaster if one of the launches failed.

The body on Earth that is not moving has acceleration zero. It seems like obvious contradiction. One force is the force that the earth exerts down on the box; that is called the weight which I will denote as W. Another force on the box is the force the table exerts up on it; that is usually called the normal force and I will denote that force as N. The only problem is that we do not know what the weight is. In order to do that, we need to devise an experiment to deduce what the magnitude of the weight is.

Unless you think the box has a different weight when it is moving than when it is not moving, there is no "obvious contradiction". But when box is at state of free fall we have the same forces: weight W and normal force N although there's no physical contact between earth and the box. Because, if we have weight that is, force in one direction we have to have opposite force with equal magnitude.

So, it is obvious contradiction because Newton is explaining two different state's free fall and body at rest with same forces. In other words, from this perspective, free fall would be impossible because acceleration would be zero. When falling, the weight is the only force on the box. It is clear that you misunderstand Newton's third law which states that "if one object exerts a force on another, the other exerts an equal and opposite force on the first. So, if the box feels a force of, say, 5 N toward the earth, the box exerts a 5 N force on the earth, toward the box.

But, in calculating the acceleration of the box, you only consider forces on the box and the only one is its weight. Once again, there is no "obvious contradiction". QUESTION: If light in earths atmosphere goes a little slower than light in a vacum, then how come when you turn on a light or laser pointer you don't hear a sonic boom? ANSWER: A sonic boom occurs when a source of sound airplane, for example, or bull whip moves through a medium faster than the speed of sound in that medium. A light source photons is not a source of sound.

The split second a bullet explodes the recoil force should start to spin the gun slightly so that the muzzle lifts up. This is because the recoil action of the gun is not directed at the center of gravity of the gun but above it. I just checked my rifle and the center of gravity is almost a inch below the center of the barrel. ANSWER: You are certainly right, an unrestrained rifle would both recoil backwards and acquire an angular velocity about the center of mass.

However, and I will not do any calculations here, because the rifle has such a large mass compared to the bullet, I am confident that the angle through which the rifle would rotate during the extremely short time the bullet is in the barrel would be trivially small. Even so, the sight mechanism is always adjusted for a certain distance and the effect you site would be included in the calibrations of the sight if it were not totally negligible. The bullet is confined to move along the barrel and the center of mass of the bullet plus gun does not move significantly during the time the bullet is in the barrel.

Therefore, I do not expect the gun to acquire an angular velocity until the bullet leaves. Again, this is contingent on the bullet having much less mass than the gun. As light is pulled toward the black hole it would accelerate, since it is already traveling at the speed of light the moment it started moving toward the black hole it would be going faster than the speed of light would it not? Just curious. What happens is that, as you would expect, it gains kinetic energy as it falls but light's energy is all kinetic. The energy of a photon is hf where h is Planck's constant and f is the corresponding frequency of the electromagnetic wave.

So, what happens when a photon gains energy is that the frequency increases; this is known as a gravitational blue shift the color of the light moves to shorter wavelengths and happens when a photon approaches any massive body, not just a black hole. I normally do all of my own research regarding how things are pronounced and such, but in this case, I'm a bit stumped.

Here's the deal: within the book, Einstein goes into a bit of math, and uses the lower case sigma which I believe means the Stefan-Boltzmann constant and the upper case sigma which I think is self-energy in this case in equations in the same section. Now, these two symbols are quite distinct when written down, but is there anyway to convey them purely verbally, or do I just need to specify "upper-case sigma" or "lower-case" sigma each time?

ANSWER: You have come to the right person since I spend several hours a week recording physics and astronomy textbooks for the blind as a volunteer. The most important part of the training is learning how to unambiguously read mathematical equations and symbols. It is conventional to read lower-case letters as just the letter and upper-case letters as "capital" or some prefer "cap". I will do one simple example. QUESTION: So, if the world were to stop spinning or even slow down in angular velocity , would the people and everything on it be considered 'lighter' or 'heavier'?

If any change were to take place? And does centrifugal force have an effect on gravity? Weight is the force which the earth exerts on you; it has nothing to do with your motion, it is a constant force which points toward the center of the earth. Hence, if you are standing on the equator, your weight is, say, lb which is about N corresponding to a mass of about 90 kg whether the earth is spinning or not. To keep things simple, let's suppose that you are standing on the equator.

Because you are moving in a circle, you are accelerating and elementary physics tells us that that acceleration is equal to the speed squared divided by the radius of the circle, about 0. Now, assuming that you know Newton's first and second laws, the sum of all the forces on you is equal to your mass times your acceleration. If you were sensitive enough, you would feel lighter. If the earth were to stop spinning, you would feel 0. If you were to stand at the north or south pole, there would no effect due to the spinning and the scale would read your weight regardless of the spin rate.

Technically, there is no such thing as centrifugal force; the force which causes you to move in a circle gravity in this case is called the centripetal force. If you want to learn more about what a centrifugal force is, see an earlier answer. Even if you do introduce a centrifugal force, it has no effect on gravity which is constant, but it has an effect on how you perceive gravity. Wouldn't that involve a force being applied, which would mean the object would lose energy to create this force?

For example, a mass hanging from a string has two forces on it, the weight down the force the earth exerts on the mass and the force upwards which the string exerts on the mass. If the mass hangs at rest, it can hang there forever without supplying any energy to the string. Energy is required only if the force acts over some distance; if that is the case, we say that the force does work and work creates or destroys energy.

The w already has g in it so we would be accounting for it twice? It depends on where you live. If your scale measures pounds, then it is measuring the force weight directly. She believed that the atom smasher could create a black hole and suck up the earth. The courts sided with the scientists. The scientists said it's not possible, and even if it did create a black hole, it would be a micro black hole and collapse in on itself. That seems to go against the laws of physics and quantum physics. We now know that matter sucked up by a black hole is permenantly imprinted on the black holes surface and the size of the black hole increases.

Did I miss something? First, some cosmic rays radiation which strikes earth from space have energies much greater than the proton energies in the LHC and if a black hole could be created and have sucked in the whole earth, that would have happened long before we evolved. The maximum kinetic energy of each proton in the collider is 7 TeV which is about equal to the mass energy of protons, so the heaviest black hole they could make would have a mass of about 14, protons, about 2. It seems to me that even if this black hole never evaporated, it could go a really long time before it got close enough to anything to suck it in.

If the tide is low, there is little to no pressure on the chain holding the buoy to the floor. But what if there is a storm and the water level rises over the buoy. Is the pressure on the chain still X or is it greater than X? Does the pressure continue to increase as the water level rises or does it stop at X?

In other words: does buoyancy increase as you dive deeper or does it have a set value? ANSWER: We do not talk about "pressure" in the chain, rather the force by the chain on what it is attached to, usually called tension; pressure is force per unit area. The buoyant force is equal to the weight of the displaced fluid. To make things clear, let's choose a specific example. Now, at low tide, attach a chain to the ocean floor which is straight but slack. As the tide rises, more and more of the buoy is under the surface.

For example, when half the volume of the buoy is submerged the buoyant force is 5 times the weight of the buoy. Now, if the water keeps rising, the same amount of the buoy is always submerged and so the tension stops increasing no matter how deep the water becomes. Although we think of water as incompressible, at extreme depths the density does get a tiny bit bigger so the buoyant force at the bottom of the ocean will be a tiny bit bigger, but a trivially small difference. The tennis balls must travel though a 5ft tube before leaving the machine. Lets say that the machine is spitting out 60 balls a second at 50 mph.

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The machine is on a swivel and a person changes the position of the machine by moving the 5ft tube left and right by pushing or pulling with his or her hand. If the speed of the balls were increased to mph would it be harder for the person to move the tube? And if so why?


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If you are interested in the mathematical details, you can link there. Now, the string breaks. A force which has no radial component the direction of the force is always perpendicular to the line drawn from the original axis of rotation to the mass is applied to keep the mass's angular velocity constant; what is the force? How is this like your question? As your ball travels down the tube, assuming the barrel has no friction, the tube exerts a force like the one described above if the tube rotates with constant angular speed.

How is it different from your question? The original question has the ball at rest in the radial direction at the beginning but your ball starts out with some radial velocity your 50 or mph. The earlier solution has two constants a and b which will be different for your problem. If you really care and are good at math, you can find the new values of the constants yourself, but I suspect you just want an answer.

Don't lose heart, this will get much less technical soon; it is just that I want to include details for my many readers who like the math! To finally answer your question, I need to put in some numbers. Since physicists do not like English units, I will use comparable metric system units:. This corresponds to about 50 0 sweep of your gun in one second.

The tube length will be 2 m about 6 ft. As you stipulated, I will choose 60 s -1 as the rate that the balls are launched, but first of all I will just look at what happens for a single ball. This is what I meant when I said things would get simpler. For all intents and purposes the force is a constant because the cosh does not vary significantly for the time the ball is in the tube.

The force you have to exert while a single fast ball is in the tube is N; the force you have to exert while a single slow ball is in the tube is 20 N. But, that is not the whole story because there may be more than one ball in the tube. The final answer: you need a force of N for the fast ball gun, N for the slow ball gun, the fast ball gun is harder to rotate. It also occurs to me that for the real-world application we should be thinking of the torque we have to apply which changes as a ball moves down the tube; oh well, I'm getting tired of this problem!

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I thought at that altitude he was in space and there was no atmosphere. Someone told me it was an effect of gravity but I just don't see that way as I think about it. Although the density of the air up there is way less than at sea level, there is still some. Standing at the front of the bus we stand in the aisle and face the back of the bus. If we throw a ball 55 mph to a person in the back of the bus what is the overall speed of the ball since it is moving in two different directions at the same speed?

I always emphasize that velocity has no meaning unless you specify velocity relative to what. In your case, somebody on the bus says the ball is going 55 mph whereas somebody by the side of the road says the ball is at rest. Those who know me say I have an attention to detail--to a fault. There is one particular element I would like to be as accurate as possible.

I'm hoping you might be able to help me. Here is the scenario: A spacecraft leaves earth on course to the moon. In order to create an Earth-like gravity inside the ship; the ship accelerates at a constant rate exerting a force on the occupants equal to one G.

Half way through the trip the craft will flip, then decelerate for the remainder of the journey. This would give the same sensation of false gravity to the occupants of the craft. So here is the question: If this were possible; how long would it take to actually reach the moon? Your scheme of having an acceleration with "constant rate" would work in empty space but not between the earth and the moon because the force causing the acceleration is not the only force on you, the earth's and moon's gravity are also acting.

So, it becomes a complicated problem as to how much force must be applied to keep the acceleration just right for where you are. The picture to the right shows you in your rocket ship. Let's call your mass M. Then there are two forces on you, your weight W down and the force the scale you are standing on exerts on you, F. W gets smaller as you get farther and farther away and you want F to always be what your weight would be on the earth's surface, Mg. Note that, for the time being, I am ignoring the moon; that would just complicate things and its force is much smaller than the earth's, at least for the first half of the trip.

I want you to understand the complication caused by the fact that W changes as you go farther away. Now, how does W change? I have plotted this in red on the graph above. The distance to the moon is about 60 earth radii. Note that for most of the trip the acceleration is just about g. I also calculated the effect the moon would have, blue dashed line, and, except for the very end of the trip, it is pretty negligible. Now that we have taken care of the always-important details, we can try to answer your question.

The symmetry of the situation is such that I need only calculate the time for the first half of the trip and double it. You can also calculate the maximum speed you would have to be about , mph halfway. I picture the force of gravity like that of a magnetic field. So I am wondering if gravity is an additive force? Like if two sphere's that were the same size and density came together, would the range of the gravitational pull increase or just in strength?

But the "range" does not change; usually, by range we mean how far the force extends and this force extends all the way to infinity. Gravity is not at all like magnetism.

You ask if it is "additive" and, yes, this force is additive but beyond your example of doubling the mass to double the force. You need not bring your two spheres together to add their forces. The way you add the forces from two different spheres is not an arithmetic sum but a vector sum, taking into account that the gravitational force from a sphere is always toward the center.

So, if you had two identical spheres separated by some distance and put another mass halfway between, it would experience zero force. They tell me I am wrong and my theory is stupid. Will a bullet fired perfectly horizontal and a bullet dropped hit the ground at the same time? My answer is yes, because gravity is constant.

The figure to the right is an actual strobed photograph of a ball red launched horizontally and another yellow dropped. If the red ball had been going faster, both would still have moved together in their vertical positions. Because a bullet has such a high velocity, air drag will have a significant effect on it but this will affect its motion in the horizontal direction, almost not at all in the vertical direction.

This is not a "theory", it is an experimental fact as the photograph shows. QUESTION: Imagine a race of beings on a distant planet observed our star and detected a small rocky planet with an oxygen rich atmosphere and decided to send an expedition to our solar system which flies in a straight line from their home world to our system at a velocity of 0. Now when the aliens are about 50LY from their destination they begin to intercept our first radio broadcasts assuming their computers can decode our signals. Then the aliens on the moving vessel record two video messages, one to transmit back to their home world saying we found life, and one to transmit forward to earth saying we found you.

Which parties would need to set their videoplayback devices to either fast forward or slow motion to observe the videos that they are receiving once the transmission got to its destination years later. Note that there is a difference between how things are and how things appear to be in relativity; see my FAQ page. QUESTION: If I take any random event that occurs on the surface of the Earth a match being struck for example , the light emitted from that event will leave the source and travel out into the universe.

If my friend could observe that event from a very distant location, say 1 LY away, he would observe it 1 year after it took place I think that is all reasonable so far, right? Now, consider if I left the Earth at something approaching the speed of light at the same time as the match was struck.

Rotational inertia

Would the observable event appear to "slow" down as I attempted to keep pace with it as it "travelled" away from the source? But if this is the case, what state would the event be in when I arrived at my friend's location 1 LY away? I mean, if it took me just over one year to get to him, he would have only just seen the event which means it must have "travelled" along with me or near me. To this end, would the event have been effectively almost frozen in time from my perspective?

But then I know that can't be the case because the speed of light is constant! That means that your clock only has ticked off 0. On your trip, you will deduce that the light gets to your friend at the end of 0. That is all as measured by a clock on your spaceship. Your friend, on the other hand, observes the event one year after it happened not 0. You cannot observe the event happening a match flaring, then dying because all that light is ahead of you. If you are interested in how things would look if you were observing the light for example if you were going toward the earth observing or going away from the earth having left before the light , see the answer just before yours.

Pls refer to the attached figure for the question that I asked. What is shown are all the forces on the block: its weight, a pulling force, a frictional force of the table horizontally, and an upward normal force from the table. You never see a reaction force if you look at forces on a body because it is not a force on the body, it is a force by the body. And, your statement that the action-reaction forces never act along the same line is totally incorrect, they always act in exactly opposite directions.

The weight is the force which the earth exerts on the block, so the reaction force is the force the block exerts on the earth which has magnitude W but points straight up. The normal force is the force which the table exerts up on the block, so the reaction force is the force the block exerts on the table which has magnitude N but points straight down. The friction is the force which the table exerts to the left on the block, so the reaction force is the force the block exerts on the table which has the same magnitude but points to the right.

The force F is the force which a string exerts on the block, so the reaction force is the force the block exerts on the earth which has magnitude F but points to the left. None of these reaction forces are shown in your free-body diagram which is as it should be because they are not forces on the block. Welcome to Newton's third law. By the way, in case you thought the equal and opposite forces W and N are an action-reaction pair, they are not; the reason they are equal and opposite is because of Newton's first law.

The question is, if I dig a hole down across the earth but avoiding the nucleus, than means instead of digging a hole straight down my feet, the hole I dig will be about 35 degrees away from the straight line across the earth. In other words, the hole will be diagonally so it will avoid the nucleus and get to the other side of the earth. If I drop a ball down the hole, will it come out from the other side? Will it get stuck in the middle since gravity pulls from the center of the earth? Does it violates the gravity law?

First, the earth's mass is uniformly distributed, that is, a cubic meter taken from anywhere has the same mass. This, of course, is not true, but it is a pretty good approximation and is the only way you can do a simple calculation of the motion in your tunnel. The motion turns out to be amazingly simple. No matter what straight line tunnel you bore through the earth, it takes the same time to travel from one end to the other, about 42 minutes. Now, I know how tension is defined, but my problem comes when trying to reason it out with force diagrams or drawings. Take this classic example: There is a tug-of-war match going on between 2 contestants.

One contestant pulls to the right with a force of N. The other pulls to the left with a force of N. What is the tension in the rope? The example proposes a situation in which the net force is zero, so there is no acceleration of the rope. The "force of tension" resists the pull from the contestant on the right with N, and the same goes for the contestant on the left. So, if we represent these forces of tension with vectors, they would both be pointing toward the "midpoint" of the rope.

That being said, why don't these tension vectors add to make N net force? Or conversely, why don't these vector forces subtract to make a 0 N net force. Also a side note: In a lot of classical mechanics problems, we always see problems involving "massless" ropes. Tension has some important properties. It is always tangent to the rope and a rope can never push, it only pulls. Of course, ropes are never massless in the real world, but we can often approximate them as if they were if they were if their mass is much less than everything else.

I can see that your main problem is that you have not been properly instructed on how to attack these simple statics problems. The first thing you have to do is to choose a body on which to focus and look only at that; you are attempting to look at the rope and both men all at once.

Let me start with a simpler example, a 1 kg mass hanging from the ceiling on a massless rope. So, I will choose a body, the 1 kg mass. What are all the forces on it? Its own weight, 10 N straight down and the force which the rope exerts up. That force by the rope is the tension. Pretty simple, the tension at the bottom of the rope must be 10 N.

Now I will choose a different body, the 1 kg mass plus half the rope. What are all the forces on the body? The weight of the body, still 10 N, and the force which the upper half of the rope exerts on the body at the point where the upper and lower halves of the rope touch. The unknown force again must be 10 N and so the tension in the middle of the rope is 10 N.

Now I will choose the 1 kg mass and the whole rope as the body. The weight of the body, still 10 N, and the force which the ceiling exerts up on the top end of the rope. So now we find that the ceiling must exert an upward force of 10 N on the top of the rope. But, Newton's third law says that if the ceiling exerts a force on the rope, the rope exerts an equal and opposite force on the ceiling. So, once again we find the tension in the rope, now at the top, is 10 N.

This is very much like your tug-o-war problem if you think about it: the weight pulls down on the rope with a force of 10 N and the ceiling pulls up on the rope with a force of 10 N, but the tension in the rope, I have just shown, is 10 N everywhere, not 20 N. Next, do the same problem but now let the mass of the rope be 1 kg also, clearly not even approximately massless. The weight of the body, now 15 N, and the force which the upper half of the rope exerts on the body at the point where the upper and lower halves of the rope touch. The unknown force must now be 15 N and so the tension in the middle of the rope is 15 N.

The weight of the body, now 20 N, and the force which the ceiling exerts up on the top end of the rope. So now we find that the ceiling must exert an upward force of 20 N on the top of the rope. So, now we find the tension in the rope at the top is 20 N. Finally, let's do your tug-o-war problem.

Part of your confusion is saying that each man exerts a force of N. It is much better to simply say that it is a tie right now so the men must be exerting equal forces in magnitude. I will choose man 1 as the body. What are all the horizontal forces on the man? The rope pulls him with an unknown force and some other force must pull him in the opposite direction so that he remains in equilibrium; that other force is simply the friction between his feet and the ground which we will take to be N. Therefore the tension at end 1 will be N.

Of course, since the rope exerts a force of N on the man, the man must be exerting a N force on the rope as was originally stated in the problem, but that is not a force on the chosen body. I will next choose man 2 as the body. Therefore the tension at end 2 will be N. I will choose man 1 plus half the rope as the body. What are all the horizontal forces on the body?

The friction pulls one direction with N and the 2 half or the rope pulls on the 1 half of the rope with the same force, so the tension in the middle of the rope is N. Suppose we choose both men plus the rope as the body. Then friction pulls to the left with N and to the right with N and the rope can be totally ignored because it does not exert any net force on the body being an internal force. I didn't understand how to apply Newton's 3rd Law to the problem. So the force of the man on the rope his pull on the rope is an action-reaction pair with the force of the rope on the man tension?

Also, do we always consider action-reaction pairs as forces that "cancel" each other out or sum to zero? Lastly, if the forces of friction between their feet and the ground are the ones that add up to a zero net force, what would happen to the tension in the case that one man was able to exert a greater force of friction in one direction? How could we calculate the tension in the rope then? There is never an occasion to add the action and reaction forces because they are forces on different things.

They do sum to zero, but that means nothing because one is on the man and one is on the rope, so why would you add them? If the men pull with different forces, the system is no longer in equilibrium and accelerates in the direction of the net force. But still, the tension in the rope is the same throughout its length.

Here is how that goes, using Newton's second law N2 it is a bit more complicated. In the picture, the man on the left in the blue shirt has a frictional force of N to the left and the man on the right in the red shirt has a frictional force N to the right.

Momentum (Stick Figure Physics Tutorials Book 3)

Each man has a mass of kg and the rope is massless. First, choose both men plus the rope as the body. The acceleration of both men will be 0. Next, choose the left man as the body. All the forces are to the left and the force the rope exerts on the blue shirt man, call it T blue. Next, choose the right man as the body. All the forces are to the right and the force the rope exerts on the red shirt man, call it T red. So, the tension in the rope is still the same everywhere you could do the above exercise of choosing one man plus half the rope to show that it is also N in the middle and, for this problem, it turns out to be just the average of the pulls at each end.

And consider that they be connected together by a very thin string. Here is the question: Can this string be gently pulled until it makes an absolutely straight line? First, suppose the string is not a string but a rigid stick but of negligible mass which we miraculously make appear between the two satellites happily orbiting in a circular orbit, one following the other. Now, we no longer have two satellites but one instead. One of the laws of classical physics says that if you want to know how a rigid body behaves under the influence of external forces, find the net force on the body and the center of mass of the body will move as if it were a point mass experiencing that force.

Now, the center of mass of the pair is below the orbit they are in and therefore this single satellite is moving too slowly to move in a circular orbit and the center of mass will change to moving in an elliptical orbit. What if you put a string of negligible mass between them in a straight line? Since it has no mass, it experiences no force and everything will behave like it was not there. Now, if you connect them with a string which has mass, it also experiences gravity and each piece of string is really wanting to be at a larger orbit so, I believe, the tendency will be for the string to curve to fit the circular orbit of the two heavier satellites.

This is an approximation assuming the string has a much smaller mass than the satellites. In reality, the string will exert forces on the satellites and alter their orbits. Under your "gently pulled" secenario, the string would pull gently also on you and the other satellite thereby altering both orbits. QUESTION: i'd like to preface my question by first stating that my knowledge of physics is "rudimentary" at it's worst, "comic book" at it's best.

Holographic universe? What does that even mean? Results indicate that students made good use of the information in the lab, and that they spent more time talking about concepts involving conservation of momentum and energy. This high-school-level assessment gauges student understanding of using diagrams to predict outcomes in colliding systems.

It is intended for use in a unit on the conservation of linear momentum. This entertaining mini-lesson developed by Exploratorium engages the learner in physical activity to explore the effects of a twisting force torque on rotational motion. By mimicking the motion of a skateboarder, students gain intuitive understanding of the law of conservation of angular momentum. This inquiry-based activity challenges students to apply their understanding of momentum and collision to determine who is at fault in an automobile accident.

It would be especially appropriate for cooperative learning groups. This PTRA manual presents the physics of impulse and momentum and learning materials to help students with this topic. Samples from the full print manual are available here. In this book, a section on theory and applications is followed by laboratory, classroom, and computer activities. The manual also includes a section on modern physics applications of the topic. Assessment questions are also provided for use by instructors.