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This is tricky! The best way I can think is to slow down time. Without resorting to high speed photography you can do this with a smooth ramp and a billiard ball. Tilt the ramp just enough so the ball will roll. Now watch the motion carefully. You could improve this experiment like Galileo did (IIRC). Add very small bumps to the ramp - fishing wire? - so ...

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If your friend agrees that before, the rock is not moving with respect to your hand and that after, the rock is moving with respect to your hand, then your friend must agree that one of the following is the case (1) the speed of the rock 'jumped' from to zero to non-zero the instant you released it or (2) the speed of the rock smoothly increased from zero ...

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Your friend is right to be skeptical. Of course, though, you are right. What he should understand is that there is definitely a force being applied to it. However, it takes time, $t$, for this force to actually change anything. 1) First make him agree that you holding the rock still makes the rock have 0 velociy or speed. 2) Then get him to understand ...

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No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ...

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There are two parts to this problem. In the first part, the stone is rising with the balloon. It has a certain acceleration for a certain time. At the end of that it will have reached a certain height and velocity. Calculate it. Then the stone is released. With the initial velocity and height calculated above, it now starts dropping. It is now subject to ...

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Someone please correct me if I'm wrong, I'm new.. The balloon is rising from the groung, so if you think about it, you have to calculate the speed the balloon reaches after 8 seconds and take it as the initial speed of the stone, then since we assume for simplicity that the air resistance is not significant, there is no other force acting on the stone, just ...

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because both objects, the earth and the moon have their own unique gravity force due to their mass., there is a place between the two where gravity works independently that keeps the separated. the moon is not falling towards the earth. actually, from some research I've done. the moon is actually moving away from the earth by 1 1/2 inches a year.. not much ...

3

While the speed and the position of an object change gradually (because of $x(t) = \int_0^t v(t') dt'$ and $v(t) = \int_0^t a(t') dt'$) the acceleration can change instantly therefore it doesn't matter for the acting force wether you just dropped the stone or not. So while you're holding the stone it feels exactly the same acceleration as you do (the force ...

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Acceleration only happens if a force is acting on an object. When you are holding the stone, the force which is accelerating the elevator is transmitted to the stone through your grip and you, the elevator and the stone form a single rigid collection of objects all accelerating at g/2. When you let go the only force left acting on the stone is gravity and ...

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While the stone is still travelling on the elevator, there are two forces acting on it, the force from the elevator to the stone, as well as the weight due to gravity. The moment the stone leaves the elevator, it becomes a free falling object. The elevator stops giving a force to the stone, and the only force remaining is its weight due to gravity. ...

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Because the initial velocity of the stone is not zero.

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ALmost freely falling . because you can't exclude the gravitational force by other planets on it . What do you think of centripetal force and falling. You are associating falling because you are very accustomed to such things but falling is solely due to an external force acted on to attract towards body so in simple circular motion when you rotate a stone ...

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The Moon is freely falling toward Earth, like you say. But it is also moving "sideways" quite quickly, so that it "misses" Earth and passes to the side. And continues to freely fall, and again misses passing to the side. Doing this in a continuous manner is called orbiting (or flying). To be a bit more technical, it is the angular momentum (and energy) of ...

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In common parlance, saying that an object is "falling" implies that the object's velocity is in the direction of the ground. However, the phrase free fall is defined as applying to an object which has no forces acting on it except for gravity, a definition that has nothing to do with velocity. If you throw a ball 3 meters above your head, your hand stops ...

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then the force you are exerting on the ball when you toss it up rules it out from free fall, right? Your hand has to be in contact with the ball to exert a force on it. As soon as you let go, you can't provide a force to the ball anymore. Since you can no longer provide a force to the ball, the only force left on it is gravity, so it's in free-fall. ...

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When the mass reaches its lowest point, the steel wire will have increased in length from $L$ to $L+x$. So equating the strain energy of the wire with the initial gravitational potential energy of the ball: $$\frac{1}{2}kx^2 = mg(L+x) \approx mgL$$ which rearranges to $$x = \sqrt{\frac{2mgL}{k}}$$ Note that $$k = \frac{EA}{L}$$ where $E$ is ...

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The difference is very minute, however the difference is there.. the difference is actually the amount that said object pulls on the Earth itself. so the difference between a 1 pound weight pulling on the Earth vs a 12 lbs bowling ball. Considering the Earth weighs in at a massive $1.317 \times 10^{25}$ lbs you could get a rough estimate by saying the Earth ...

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