The mass of the person would be 66 kg on both Earth and the Moon.
Answer:304 kg*ms or in ms 84.48 kgm/s
Explanation:
Answer:
Hey There!
Let's solve...
I think your diagram must be similar to whatever i drew on attachment...
Now.....
![n_{1} = ma_{1} \\ \\ mg - f_{1} - t = 2ma_{1} \\ \\ mg - \mu _{1} ma_{1} - t = 2ma_{1} \\ \\ t = mg - ma_{1}( \mu_{1} + 2 )](https://tex.z-dn.net/?f=%20n_%7B1%7D%20%3D%20%20ma_%7B1%7D%20%5C%5C%20%20%5C%5C%20mg%20-%20%20f_%7B1%7D%20-%20t%20%3D%20%202ma_%7B1%7D%20%5C%5C%20%20%5C%5C%20mg%20-%20%20%5Cmu%20_%7B1%7D%20ma_%7B1%7D%20-%20t%20%3D%20%202ma_%7B1%7D%20%5C%5C%20%20%5C%5C%20t%20%3D%20mg%20%20-%20%20%20ma_%7B1%7D%28%20%20%5Cmu_%7B1%7D%20%2B%202%20%29%20%20)
Lets solve bigger block which is along vertical
![n_{2} = mg + f_{1} + t \\ \\ n_{2} = mg + \mu_{1} ma_{1} + mg - \\ ma_{1} ( \mu_{1} + 2 ) \\ \\ n_{2} = (m + m)g - 2ma_{1} \\ \\](https://tex.z-dn.net/?f=%20n_%7B2%7D%20%3D%20mg%20%2B%20%20f_%7B1%7D%20%2B%20t%20%5C%5C%20%20%5C%5C%20%20n_%7B2%7D%20%3D%20mg%20%2B%20%20%20%5Cmu_%7B1%7D%20ma_%7B1%7D%20%2B%20mg%20-%20%20%5C%5C%20%20ma_%7B1%7D%20%28%20%20%5Cmu_%7B1%7D%20%2B%202%20%29%20%20%5C%5C%20%20%5C%5C%20%20n_%7B2%7D%20%3D%20%28m%20%2B%20m%29g%20-%20%202ma_%7B1%7D%20%5C%5C%20%20%5C%5C%20%20%20%20%20%20%20)
Lets solve it horizontally
![2t - n_{1} - f_{2} = ma_{1} \\ \\ 2mg - 2ma_{1}( \mu_{1} + 2) - ma_{1} \\ - \mu_{2} n_{2} = ma_{1}](https://tex.z-dn.net/?f=2t%20-%20%20n_%7B1%7D%20-%20%20f_%7B2%7D%20%3D%20%20ma_%7B1%7D%20%5C%5C%20%20%5C%5C%202mg%20-%20%202ma_%7B1%7D%28%20%20%5Cmu_%7B1%7D%20%2B%202%29%20%20-%20%20ma_%7B1%7D%20%5C%5C%20%20-%20%20%20%5Cmu_%7B2%7D%20n_%7B2%7D%20%3D%20%20ma_%7B1%7D%20%20%20)
![2mg - \mu_{2} n_{2} = a_{1}(m + m + 2m\mu_{1} + 4_{m})](https://tex.z-dn.net/?f=2mg%20-%20%20%20%5Cmu_%7B2%7D%20n_%7B2%7D%20%3D%20%20a_%7B1%7D%28m%20%2B%20m%20%2B%20%202m%5Cmu_%7B1%7D%20%2B%20%204_%7Bm%7D%29%20%20)
![2mg - \mu_{2}^{(m + m)}g + \mu_{2}2ma_{1} \\ \\ = a_{1}(m + 5m + 2 \mu_{1}) \\ \\ g(2m - \mu_{2}m - \mu_{2}m) = \\ \\ a_{1}(m + 5m + 2 \mu_{1} - 2 \mu_{2})](https://tex.z-dn.net/?f=2mg%20-%20%20%20%5Cmu_%7B2%7D%5E%7B%28m%20%2B%20m%29%7Dg%20%2B%20%20%5Cmu_%7B2%7D2ma_%7B1%7D%20%5C%5C%20%20%5C%5C%20%20%3D%20%20a_%7B1%7D%28m%20%2B%205m%20%2B%202%20%5Cmu_%7B1%7D%29%20%5C%5C%20%20%5C%5C%20g%282m%20-%20%20%5Cmu_%7B2%7Dm%20-%20%20%20%5Cmu_%7B2%7Dm%29%20%3D%20%20%5C%5C%20%20%5C%5C%20%20a_%7B1%7D%28m%20%2B%205m%20%2B%20%202%20%5Cmu_%7B1%7D%20-%20%202%20%5Cmu_%7B2%7D%29%20%20%20%20%20%20%20%20)
Final answer is just make fraction of that
![a_{1} = \frac{g(2m - \mu _{2}m - \mu_{2}m )}{(m + 5m + 2m \mu_{1} - 2m \mu_{2})}](https://tex.z-dn.net/?f=%20a_%7B1%7D%20%3D%20%20%5Cfrac%7Bg%282m%20-%20%20%20%5Cmu%20_%7B2%7Dm%20-%20%20%20%20%20%5Cmu_%7B2%7Dm%20%29%7D%7B%28m%20%2B%205m%20%2B%202m%20%5Cmu_%7B1%7D%20-%202m%20%5Cmu_%7B2%7D%29%7D%20)
so it is your answer.....
<h3>I hope it is helpful to you....</h3><h3>Cheers!___________</h3>
Answer: B
Explanation:
Sound and vibrations travel faster through solids than gases.
Answer:
The statement is true but does not fully explain the situation.
Explanation:
- According to Newton's third law, a pair of equal and opposite forces is generated in every action and they of course cancel out when the system (the doer and the receiver are taken as a single system).
- According to the same law, one can work out the force-pairs that are generated during the push on the truck as follows.
- The friction force that applies to the truck is equal to the force applied to it by the person as the truck does not move.
- It is this friction force that keeps the truck stationary on the ground.
- To move the truck one should apply a force that is just above the static friction force that could apply to the truck from the ground at a given instance.
- Since the truck is massive, the static friction/ the maximum friction is also massive which a person could not imagine even applying.
- And that is why a push by a person does not move a massive truck.
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