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lara [203]
3 years ago
10

A certain string can withstand a maximum tension of 39. N without breaking. A child ties a 0.43 kg stone to one end and, holding

the other end, whirls the stone in a vertical circle of radius 0.98 m, slowly increasing the speed until the string breaks. Where is the stone on its path when the string breaks (top of the circle, middle of the circle or bottom of the circle)
Physics
1 answer:
frez [133]3 years ago
6 0

Answer:

Bottom of the circle.

Explanation:

At the top of the circle the tension and the weight contribute on being the centripetal force, at the middle of the circle only the tension contributes on being the centripetal force (the weight being perpendicular to it), while <u>at the bottom</u> of the circle the tension contributes on being the centripetal force (as always) <em>but the weight against to it</em>, so here is where the tension must be greater to allow the same centripetal force as the other cases, thus here is where the string will break.

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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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\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

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