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marusya05 [52]
2 years ago
8

75 points

Physics
1 answer:
Vikki [24]2 years ago
5 0

Answer:

Explanation:

The height to which a ball will bounce depends on the height from which it is dropped, what the ball is made out of (and if it is inflated, what the pressure is), and what the surface it bounces from is made out of. The radius of the ball doesn't really matter, if you are measuring the height of the ball from the bottom of the ball to the ground.

A ball's gravitational potential energy is proportional to its height. At the bottom, just before the bounce, this energy is now all in the form of kinetic energy. After the bounce, the ball and the ground or floor have absorbed some of that energy and have become warmer and have made a noise. This energy lost in the bounce is a more or less constant fraction of the energy of the ball before the bounce. As the ball goes back up, kinetic energy (now a bit less) gets traded back for gravitational potential energy, and it will rise back to a height that is the original height times (1-fraction of energy lost). We'll call this number f. For a superball, f may be around 90% (0.9) or perhaps even bigger. For a steel ball on a thick steel plate, f is >0.95. For a properly inflated basketball, f is about 0.75. For a squash ball, f might be less than 0.5 or 0.25 - squash balls are not very bouncy. The steel ball on an unvarnished pine wood floor may not bounce at all, but rather make a dent, and so what the floor is made out of makes quite a lot of difference.

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Ray Of Light [21]

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about 10 ms^-^2, the density of whatever you're sinking in, and the depth at which you are. In formula, p(h) = p_0 + \rho g h, and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - 1000 * 10 * 1 = 10^4 Pa, while in B is 1000*10*2 = 2*10^4 Pa, so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is 800*10*2 = 1,6*10^4 Pa: Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so 4m*3m*2m = 24m^3

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: 800* 24 = 1,92 *10^4kg

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's 1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N

2d. \rho gh again, what a surprise! 800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa

3. Yet again, \rho gh. 1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa

4 0
2 years ago
A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds
Inessa05 [86]

Answer:

A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel

20miles / 5sec = 4miles /sec would be the average speed for the last 20 m

Explanation:

The answer is 4 m/s.

In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).

The relation of speed (v), distance (d), and time (t) can be expressed as:

v = d/t

We need to calculate the speed of the second 5 seconds of the travel:

d = 20 m (total 30 meters - first 10 meters)

t = 5 s (time from t = 5 seconds to t = 10 seconds)

Thus:

v = 20m / 5s = 4 m/s

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aleksandr82 [10.1K]

Answer:

Period is 86811.5 seconds.

Explanation:

{ \boxed{ \bf{T {}^{2} =  (\frac{4 {\pi}^{2} }{GM}) {r}^{3}   }}}

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Answer:

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