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natali 33 [55]
3 years ago
12

A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p

ositive test charge (q0 = +4.56 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -6.95 x 10-9 J. Find rB.
Physics
1 answer:
ella [17]3 years ago
7 0

Answer:

r_B = 1.88 m

Explanation:

As we know that work done by electric force is given as

W_e = -q\Delta V

so here we know that charge is moving from

r_A = 2.97 m

to another position

so we will have

W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}

-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})

-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})

r_B = 1.88 m

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A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length
Ivahew [28]

Answer:

<em>The total length of the spring would be 0.65 m</em>

Explanation:

The Concept

Hooke's law evaluates the increment of  spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the  force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;

F = k x ..................1

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Step by Step Calculations

We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;

x = F/k

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x = 0.14715 m

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L = 0.14715 m + 0.5 m = 0.64715

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Answer:

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Explanation:

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To the nearest degree,

θ2 = 20 degrees.

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