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OLga [1]
3 years ago
14

A charge of 19 nC is uniformly distributed along a straight rod of length 15 m that is bent into a circular arc with a radius of

5.7 m. What is the magnitude of the electric field at the center of curvature of the arc?

Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

Answer:

The magnitude of the electric field at the center of curvature of the arc is 3.87 N/C

Explanation:

Please see the attachments below

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a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
Mademuasel [1]
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

And then, substituting T2 into (1), we find
T_1 = 1717 N
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3 years ago
Given the diagram below determine the net force acting on the object
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A) 0

because all of the forces cancel out, so it is not moving with balanced forces.

7 0
3 years ago
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Choose all the answers that apply.
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<span>According to Newton's first law of motion:

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</span>
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3 years ago
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