Jill is the input, as she creates the force. The wrench is the output because it gives the force to the finish peace of the chain.
The elephant and the mouse having zero weight in a gravity free space will not bump into you at the same effect.
<u>Explanation:
</u>
When both are in a gravity free space, the weights are zero, as we know that the![\text {weight of the body}=\text {mass of the body} \times \text {acceleration due to gravity}](https://tex.z-dn.net/?f=%5Ctext%20%7Bweight%20of%20the%20body%7D%3D%5Ctext%20%7Bmass%20of%20the%20body%7D%20%5Ctimes%20%5Ctext%20%7Bacceleration%20due%20to%20gravity%7D)
![\text {here, the weight of elephant}=\text {mass of elephant } \times \text {zero gravti} y=zero](https://tex.z-dn.net/?f=%5Ctext%20%7Bhere%2C%20the%20weight%20of%20elephant%7D%3D%5Ctext%20%7Bmass%20of%20elephant%20%7D%20%5Ctimes%20%5Ctext%20%7Bzero%20gravti%7D%20y%3Dzero)
![\text {similarly,weight of mouse}=\text {mass of mouse } \times \text {zero gravity}=zero](https://tex.z-dn.net/?f=%5Ctext%20%7Bsimilarly%2Cweight%20of%20mouse%7D%3D%5Ctext%20%7Bmass%20of%20mouse%20%7D%20%5Ctimes%20%5Ctext%20%7Bzero%20gravity%7D%3Dzero)
But when they will acquire the speed of same magnitude, say v, their different masses will acquire different momentum, which will make the difference in effect while bumping.
![\text { momentum of mouse = mass of mouse } \times v](https://tex.z-dn.net/?f=%5Ctext%20%7B%20momentum%20of%20mouse%20%3D%20mass%20of%20mouse%20%7D%20%5Ctimes%20v)
And as we know
Therefore, effect of impact by elephant will be more than that of mouse
. An elephant breaking into you will take you back faster than a mouse in space hits you.
Answer:
a)0.5564c
b)43.6 m
Explanation:
Given proper length of falcrum L₀= 43.6 m
improper length L=30.1 m (when viewed from moving frame)
we know that ![L=L_{0}\sqrt{1-\frac{v^2}{c^2}}](https://tex.z-dn.net/?f=L%3DL_%7B0%7D%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D)
![30.1=43.6\sqrt{1-\frac{v^2}{c^2}](https://tex.z-dn.net/?f=30.1%3D43.6%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D)
⇒
=![\frac{30.1}{43.6}](https://tex.z-dn.net/?f=%5Cfrac%7B30.1%7D%7B43.6%7D)
![{1-\frac{v^2}{c^2}}=0.6904](https://tex.z-dn.net/?f=%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%3D0.6904)
![\frac{v^2}{c^2}=0.3096](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%3D0.3096)
![v^{2}=3096c^{2}](https://tex.z-dn.net/?f=v%5E%7B2%7D%3D3096c%5E%7B2%7D)
v=0.5564c
this is the required speed of falcon when it passes luke
b). Since Han solo is on the Falcon its reference frame will be falcon itself hence there wont be any change in the length of Falcon that its length will be
43.6 m
Soory i think multiple of velocity and angle