The answer is: [C]: "elasticity" .
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Answer:
Differences between freefall and weightlessness are as follows:
<h3>
<u>Freefall</u></h3>
- When a body falls only under the influence of gravity, it is called free fall.
- Freefall is not possible in absence of gravity.
- A body falling in a vacuum is an example of free fall.
<h3>
<u>Weightlessness</u></h3>
- Weightlessness is a condition at which the apparent weight of body becomes zero.
- Weightlessness is possible in absence of gravity.
- A man in a free falling lift is an example of weightlessness.
Hope this helps....
Good luck on your assignment....
1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:
![F=G\frac{Mm}{r^2}](https://tex.z-dn.net/?f=F%3DG%5Cfrac%7BMm%7D%7Br%5E2%7D)
where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find
![F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N](https://tex.z-dn.net/?f=F%3D%286.67%5Ccdot%2010%5E%7B-11%7Dm%5E3%20kg%5E%7B-1%7D%20s%5E%7B-2%7D%29%5Cfrac%7B%288.7%5Ccdot%2010%5E%7B13%7D%20kg%29%28110%20kg%29%7D%7B%282000%20m%29%5E2%7D%3D0.16%20N)
2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force
![m\frac{v^2}{r}=G\frac{Mm}{r^2}](https://tex.z-dn.net/?f=m%5Cfrac%7Bv%5E2%7D%7Br%7D%3DG%5Cfrac%7BMm%7D%7Br%5E2%7D)
where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:
![v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7BGM%7D%7Br%7D%7D%3D%5Csqrt%7B%5Cfrac%7B%286.67%5Ccdot%2010%5E%7B-11%7D%20m%5E3%20kg%5E%7B-1%7D%20s%5E%7B-2%7D%29%288.7%5Ccdot%2010%5E%7B13%7D%20kg%29%7D%7B2000%20m%7D%7D%3D1.7%20m%2Fs)
Initial velocity = ![\(v_0\)](https://tex.z-dn.net/?f=%5C%28v_0%5C%29)
acceleration in the downward direction = -9.8 ![\(\frac {m}{s^2}\)](https://tex.z-dn.net/?f=%5C%28%5Cfrac%20%7Bm%7D%7Bs%5E2%7D%5C%29)
Final velocity at the highest point = 0
Maximum height reached = 0.410 m
Now, Using third equation of motion:
![\(v^2 = {v_0}^{2} + 2aH](https://tex.z-dn.net/?f=%5C%28v%5E2%20%3D%20%7Bv_0%7D%5E%7B2%7D%20%2B%202aH)
![\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410](https://tex.z-dn.net/?f=%5C%280%5E2%20%3D%20%7Bv_0%7D%5E%7B2%7D%20-%202%20%5Ctimes%209.8%20%5Ctimes%200.410)
![\({v_0}^{2} = 2 \times 9.8 \times 0.410\)](https://tex.z-dn.net/?f=%5C%28%7Bv_0%7D%5E%7B2%7D%20%3D%202%20%5Ctimes%209.8%20%5Ctimes%200.410%5C%29)
![\(v_0 = 2.834 \frac {m}{s}\)](https://tex.z-dn.net/?f=%5C%28v_0%20%3D%202.834%20%5Cfrac%20%7Bm%7D%7Bs%7D%5C%29)
Speed with which the flea jumps = ![\(2.834 \frac {m}{s}\)](https://tex.z-dn.net/?f=%5C%282.834%20%5Cfrac%20%7Bm%7D%7Bs%7D%5C%29)
Explanation:
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