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disa [49]
3 years ago
6

A ball of mass 5 kg attached to a string is swung in a horizontal circle of radius 0.5 m. If the tension in the string is 10 N,

the work done by the tension in one revolution is _____________ J.
Physics
1 answer:
kirill115 [55]3 years ago
8 0

Answer:

0 J

Explanation:

given,

mass of the ball = 5 kg

radius of the horizontal circle = 0.5 m

tension in the string = 10 N

Work done = ?

Work done by the tension in the circular path will be equal to zero.

This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.

so, work done = F s cos θ

     θ = 90°,

work done = F s cos 90°        ∵ cos 90° = 0

Work done = 0 J

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Answer:

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Explanation:

Given;

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V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 =  \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

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