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

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>
- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
brainly.com/question/28045397
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I believe the answer is the mass of the object and the speed at which it is moving.
In the question "Which of the following is a common human relations mistake?" The correct answer is "Misjudging others' abilities<span>"
The other answer options "Saying thank you when somebody does something nice" and "Taking responsibility for mistakes" are all good human realation practices.</span>
The slowest line is the solid line and the fastest is the dotted line that crosses the solid line
for future reference you just need to find the slope or the line which is traveling most vertical