Answer:
I think C
Explanation:
Since the bus is moving away from John.
{C - V}.
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass 
kg, charge +e = 
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 
m/s. The proton comes momentarily to rest at a distance 
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 
m apart?
Explanation:
The given data is as follows.
Mass of proton = 
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
= 
where, 
U = 
Putting the given values into the above formula as follows.
U =
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .
Answer:
F₃ = 122.88 N
θ₃ = 20.63°
Explanation:
First we find the components of F₁:
For x-component:
F₁ₓ = F₁ Cos θ₁
F₁ₓ = (50 N) Cos 60°
F₁ₓ = 25 N
For y-component:
F₁y = F₁ Sin θ₁
F₁y = (50 N) Sin 60°
F₁y = 43.3 N
Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:
F₂ₓ = F₂ = 90 N
F₂y = 0 N
Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:
F₁ₓ + F₂ₓ + F₃ₓ = 0 N
25 N + 90 N + F₃ₓ = 0 N
F₃ₓ = - 115 N
for y-components:
F₁y + F₂y + F₃y = 0 N
43.3 N + 0 N + F₃y = 0 N
F₃y = - 43.3 N
Now, the magnitude of F₃ can be found as:
F₃ = √F₃ₓ² + F₃y²
F₃ = √[(- 115 N)² + (- 43.3 N)²]
<u>F₃ = 122.88 N</u>
and the direction is given as:
θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)
<u>θ₃ = 20.63°</u>
I think its all four of them could be wrong but try all four !!!!!!
