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3 years ago

11

A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second,

what is the tension in the string?

1 answer:

Aloiza [94]3 years ago

4 0

Answer:

19.74 N

Explanation:

mass of ball (m) = 0.25 kg

radius (r) = 0.5 m

time (t) = 2 revolutions per seconds = 1/2 = 0.5 second per revolution

find the tension in the string

tension (T) = $\frac{mv^{2} }{r}$

where velocity (v) = $\frac{2πr}{t}$

tension now becomes (T) = $\frac{m}{r} x (\frac{2πr}{t})^{2}$

tension (T) = $\frac{4π^{2}rm }{t^{2} }$

now substituting the values of mass (m), time (t) and radius (r) into the equation above we have

tension (T) = $\frac{4π^{2}x0.5x0.25}{0.5^{2} }$

tension (T) = $2π^{2}$ = $2x3.142^{2}$ = 19.74 N

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On that line segment between the two charges, at approximately $0.7\; \rm m$ away from the smaller charge (the one with a magnitude of $5 \times 10^{-19}\; \rm C$ ,) and approximately $1.3\; \rm m$ from the larger charge (the one with a magnitude of $20 \times 10^{-19}\; \rm C$ .)

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Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

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The electric field due to $q_1 = 5\times 10^{-19}\; \rm C$ would have a magnitude of:

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