the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or 
ATQ,
We have density of charge,
λ = 
Where L is the rod's length, in this case the semicircle's length L = πr
Q is the charge on the rod
The potential created at the center by an differential element of charge is:
where k is the coulomb's constant
r is the distance from dQ to center of the circle
v = ∫ 
, Where a = radius, k = 1 / 4πε∘
v = or Q ∕ (4πε∘a)
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The displacement of the train after 2.23 seconds is 25.4 m.
<h3>Resultant velocity of the train</h3>
The resultant velocity of the train is calculated as follows;
R² = vi² + vf² - 2vivf cos(θ)
where;
- θ is the angle between the velocity = (90 - 51) + 37 = 76⁰
R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)
R² = 129.75
R = √129.75
R = 11.39 m/s
<h3>Displacement of the train</h3>
The displacement of the train is the change in position of the train after a given period of time.
The displacement is calculated as follows;
Δx = vt
Δx = 11.39 m/s x 2.23 s
Δx = 25.4 m
Thus, the displacement of the train after 2.23 seconds is 25.4 m.
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Answer: 80 N downward
Explanation:
705 - 625 = 80 N downward
Answer:
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Explanation:
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