Answer:
Total moment of inertia when arms are extended: 1.613 
Explanation:
This second part of the problem could be a pretty complex one, but if they expect you to do a simple calculation, which is what I imagine, the idea is just adding another moment of inertia to the first one due to the arms extended laterally and use the moment of inertia for such as depicted in the image I am attaching.
In that image:
L is the length from one end to the other of the extended arms (each 0.75m from the center of the body) which gives 1.5 meters.
m is the mass of both arms. That is: twice 5% of the mass of the person: which mathematically can be written as: 2 * 0.05 * 56.5 = 5.65 kg
Therefore this moment of inertia to be added can be obtained using the formula shown in the image:
Now, one needs to add this to the previous moment that you calculated, resulting in:
0.554 + 1.059 = 1.613
The Bohr model of atom addressed the problem of why electrons did not fall into the positive nucleus.
As per Rutherford's atomic model, electrons are revolving around the nucleus just like planets are revolving around the sun. But his description about electronic motion was against the Clark Maxwell's electromagnetic theory .
As per Maxwell any charged particle under accelerated motion will release energy. The electron is a negatively charged particle which under acceleration will come closer and closer to the nucleus and finally it will fall on the nucleus for which the whole atom will be destroyed. The path of electron will be a spiral one instead of circular. Hence Rutherford's theory could not explain the stability of nucleus.
Bohr addressed that electron in particular orbit has a fixed amount of energy .There will be no emission or absorption of energy as long as electron revolves around the nucleus.The electron will absorb energy when it will jump from ground state to excited state and will emit energy when it will jump from excited state to ground state.
Hence the option A is right,.
A black hole is a place in space where gravity pulls so much that even light can not get out.
Only the horizontal component of the tension force is performing work on the crate. This component has magnitude (50 N) cos(30°) = 25√3 N, so the work done is
(25√3 N) (10 m) = 250√3 N•m ≈ 433 J
