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2

For the record, here's a worked solution: If $r$ is the distance between the two point masses $m_1$ and $m_2$--which start at rest--then as both accelerate towards each other, where $$\frac{d^2r}{dt^2} = -\frac{Gm}{r^2} \ \ \ , \ \ \ \ \ m = m_1 + m_2$$ The first step in solving this equation is the least obvious: Multiply both sides by $\displaystyle ... 0 This is tricky! The best way I can think is to slow down time. Without resorting to high speed photography you can do this with a smooth ramp and a billiard ball. Tilt the ramp just enough so the ball will roll. Now watch the motion carefully. You could improve this experiment like Galileo did (IIRC). Add very small bumps to the ramp - fishing wire? - so ... 1 If your friend agrees that before, the rock is not moving with respect to your hand and that after, the rock is moving with respect to your hand, then your friend must agree that one of the following is the case (1) the speed of the rock 'jumped' from to zero to non-zero the instant you released it or (2) the speed of the rock smoothly increased from zero ... 2 Your friend is right to be skeptical. Of course, though, you are right. What he should understand is that there is definitely a force being applied to it. However, it takes time,$t$, for this force to actually change anything. 1) First make him agree that you holding the rock still makes the rock have 0 velociy or speed. 2) Then get him to understand ... 24 No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ... 3 There are two parts to this problem. In the first part, the stone is rising with the balloon. It has a certain acceleration for a certain time. At the end of that it will have reached a certain height and velocity. Calculate it. Then the stone is released. With the initial velocity and height calculated above, it now starts dropping. It is now subject to ... -1 Someone please correct me if I'm wrong, I'm new.. The balloon is rising from the groung, so if you think about it, you have to calculate the speed the balloon reaches after 8 seconds and take it as the initial speed of the stone, then since we assume for simplicity that the air resistance is not significant, there is no other force acting on the stone, just ... 0 because both objects, the earth and the moon have their own unique gravity force due to their mass., there is a place between the two where gravity works independently that keeps the separated. the moon is not falling towards the earth. actually, from some research I've done. the moon is actually moving away from the earth by 1 1/2 inches a year.. not much ... 3 While the speed and the position of an object change gradually (because of$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 ...

4

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 ...

11

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. ...

0

Because the initial velocity of the stone is not zero.

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