Answer:
Explanation:
Apply the law of conservation of energy

![Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)](https://tex.z-dn.net/?f=Gm_1m_2%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1v_1%5E2%2Bm_2v_2%5E2%29)
from the law of conservation of the linear momentum

Therefore,
![Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)](https://tex.z-dn.net/?f=Gm_1m_2%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1v_1%5E2%2Bm_2v_2%5E2%29)
![=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%5Bm_1v_1%5E2%2Bm_2%5B%5Cfrac%7Bm_1v_1%7D%7Bm_2%7D%20%5D%5E2%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%5Bm_1v_1%5E2%2B%5Cfrac%7Bm_1%5E2v_1%5E2%7D%7Bm_2%7D%20%5D%5C%5C%5C%5C%3D%5Cfrac%7Bm_1v_1%5E2%7D%7B2%7D%20%5B%5Cfrac%7Bm_1%2Bm_2%7D%7Bm_2%7D%20%5D)
![v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]](https://tex.z-dn.net/?f=v_1%5E2%3D%5B%5Cfrac%7B2Gm_2%5E2%7D%7Bm_1%2Bm_2%7D%20%5D%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D)
Substitute the values in the above result
![v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]](https://tex.z-dn.net/?f=v_1%5E2%3D%5B%5Cfrac%7B2Gm_2%5E2%7D%7Bm_1%2Bm_2%7D%20%5D%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D)
![=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B2%286.67%5Ctimes%2010%5E-%5E1%5E1%29%28107%29%5E2%7D%7B27%2B107%7D%20%5D%5B%5Cfrac%7B1%7D%7B26%7D%20-%5Cfrac%7B1%7D%7B41%7D%5D%20%5C%5C%5C%5C%3D1.6038%5Ctimes%2010%5E-%5E1%5E0%5C%5C%5C%5Cv_1%3D%5Csqrt%7B1.6038%5Ctimes%20106-%5E1%5E0%7D%20%5C%5C%5C%5C%3D1.2664%20%5Ctimes%2010%5E-%5E5m%2Fs)
B) the speed of the sphere with mass 107.0 kg is

\\\\=3.195\times 10^-^6m/s](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B27%7D%7B107%7D%20%5D%281.2664%20%5Ctimes%2010%5E-%5E5%29%5C%5C%5C%5C%3D3.195%5Ctimes%2010%5E-%5E6m%2Fs)
C) the magnitude of the relative velocity with which one sphere is

D) the distance of the centre is proportional to the acceleration

Thus,

and

When the sphere make contact with eachother
Therefore,

And

The point of contact of the sphere is

The answer is 100mm/s. I hope this helps :)
An alcohol functional group is called a hydroxyl group.
Your answer is 2.5 x 10^(11).
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On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons
BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N