Kinetic Energy is movement energy (most simplistic way I can put it) so its motion.
The correct answer to the question is : D) Be moving at a constant velocity.
EXPLANATION:
As per Newton's first laws of motion, every body continues to be at state of rest or of uniform motion in a straight line unless and until it is compelled by some external unbalanced forces acting on it.
Hence, it is the unbalanced force which changes the state of rest or motion of a body. Balanced force is responsible for keeping the body to be either in static equilibrium or in dynamic equilibrium.
As per the options given in the question, the last one is true for an object under balanced forces.
It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just
Answer:
The angular velocity is ![w_f = 1.531 \ rad/ s](https://tex.z-dn.net/?f=w_f%20%3D%20%201.531%20%5C%20rad%2F%20s)
Explanation:
From the question we are told that
The mass of each astronauts is ![m = 50 \ kg](https://tex.z-dn.net/?f=m%20%3D%20%2050%20%5C%20kg)
The initial distance between the two astronauts ![d_i = 7 \ m](https://tex.z-dn.net/?f=d_i%20%20%3D%20%207%20%5C%20%20m)
Generally the radius is mathematically represented as ![r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m](https://tex.z-dn.net/?f=r_i%20%20%3D%20%20%5Cfrac%7Bd_i%7D%7B2%7D%20%3D%20%5Cfrac%7B7%7D%7B2%7D%20%20%3D%20%203.5%20%5C%20%20m)
The initial angular velocity is ![w_1 = 0.5 \ rad /s](https://tex.z-dn.net/?f=w_1%20%3D%200.5%20%5C%20%20rad%20%2Fs)
The distance between the two astronauts after the rope is pulled is ![d_f = 4 \ m](https://tex.z-dn.net/?f=d_f%20%3D%20%204%20%5C%20%20m)
Generally the radius is mathematically represented as ![r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m](https://tex.z-dn.net/?f=r_f%20%20%3D%20%20%5Cfrac%7Bd_f%7D%7B2%7D%20%3D%20%5Cfrac%7B4%7D%7B2%7D%20%20%3D%20%202%5C%20%20m)
Generally from the law of angular momentum conservation we have that
![I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}](https://tex.z-dn.net/?f=I_%7Bk_1%7D%20w_%7Bk_1%7D%2B%20I_%7Bp_1%7D%20w_%7Bp_1%7D%20%3D%20I_%7Bk_2%7D%20w_%7Bk_2%7D%2B%20I_%7Bp_2%7D%20w_%7Bp_2%7D)
Here
is the initial moment of inertia of the first astronauts which is equal to
the initial moment of inertia of the second astronauts So
![I_{k_1} = I_{p_1 } = m * r_i^2](https://tex.z-dn.net/?f=I_%7Bk_1%7D%20%3D%20I_%7Bp_1%20%7D%20%3D%20%20m%20%2A%20%20r_i%5E2)
Also
is the initial angular velocity of the first astronauts which is equal to
the initial angular velocity of the second astronauts So
![w_{k_1} =w_{p_1 } = w_1](https://tex.z-dn.net/?f=w_%7Bk_1%7D%20%3Dw_%7Bp_1%20%7D%20%3D%20w_1)
Here
is the final moment of inertia of the first astronauts which is equal to
the final moment of inertia of the second astronauts So
![I_{k_2} = I_{p_2} = m * r_f^2](https://tex.z-dn.net/?f=I_%7Bk_2%7D%20%3D%20I_%7Bp_2%7D%20%3D%20%20m%20%2A%20%20r_f%5E2)
Also
is the final angular velocity of the first astronauts which is equal to
the final angular velocity of the second astronauts So
![w_{k_2} =w_{p_2 } = w_2](https://tex.z-dn.net/?f=w_%7Bk_2%7D%20%3Dw_%7Bp_2%20%7D%20%3D%20w_2)
So
![mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2](https://tex.z-dn.net/?f=mr_i%5E2%20w_1%20%2B%20mr_i%5E2%20w_1%20%3D%20mr_f%5E2%20w_2%20%2B%20mr_f%5E2%20w_2)
=> ![2 mr_i^2 w_1 = 2 mr_f^2 w_2](https://tex.z-dn.net/?f=2%20mr_i%5E2%20w_1%20%3D%202%20mr_f%5E2%20w_2)
=> ![w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }](https://tex.z-dn.net/?f=w_f%20%3D%20%20%5Cfrac%7B2%20%2A%20m%20%2A%20r_i%5E2%20w_1%7D%7B2%20%2A%20m%20%2A%20%20r_f%5E2%20%7D)
=> ![w_f = \frac{3.5^2 * 0.5}{ 2^2 }](https://tex.z-dn.net/?f=w_f%20%3D%20%20%5Cfrac%7B3.5%5E2%20%2A%20%200.5%7D%7B%20%202%5E2%20%7D)
=> ![w_f = 1.531 \ rad/ s](https://tex.z-dn.net/?f=w_f%20%3D%20%201.531%20%5C%20rad%2F%20s)
Answer:
R = 715.4 N
L = 166.6 N
Explanation:
ASSUME the painter is standing right of center
Let L be the left rope tension
Let R be the right rope tension
Sum moments about the left end to zero. Assume CCW moment is positive
R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0
R = 715.4 N
Sum moments about the right end to zero
20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0
L = 166.6 N
We can verify by summing vertical forces
116.6 + 715.4 - (70 + 20)(9.8) ?=? 0
0 = 0 checks
If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.