This question involves the concepts of the law of conservation of momentum and velocity.
The velocity of the eight ball is "5.7 m/s".
According to the law of conservation of momentum:

where,
m₁ = mass of number three ball = 5 g
m₂ = mass of the eight ball = 6 g
u₁ = velocity of the number three ball = 3 m/s
u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)
v₁ = final velocity of the three number ball = - 5 m/s
v₂ = final velocity of the eight ball = ?
Therefore,
(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)

<u>v₂ = 5.7 m/s</u>
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Answer:
A
Explanation:
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The correct option is A.
Answer:
The moment of inertia is 
Explanation:
The moment of inertia is equal:

If r is 
and 


If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Given the data in the question;
- Length of the massless beam;

- Distance of support from the left end;

- First mass;

- Distance of beam from the left end( m₁ is attached to );

- Second mass;

- Distance of beam from the right of the support( m₂ is attached to );

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.
Hence, 
we divide both sides by 

Next, we make
, the subject of the formula
![x_1 = x - [ \frac{m_2x_2}{m_1} ]](https://tex.z-dn.net/?f=x_1%20%3D%20x%20-%20%5B%20%5Cfrac%7Bm_2x_2%7D%7Bm_1%7D%20%5D)
We substitute in our given values
![x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]](https://tex.z-dn.net/?f=x_1%20%3D%203.00m%20-%20%5B%20%5Cfrac%7B61.7kg%5C%20%2A%20%5C%200.273m%7D%7B31.3kg%7D%20%5D)


Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Learn more; brainly.com/question/3882839
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