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Alex17521 [72]
2 years ago
14

How fast do you need to swing a 250-g ball at the end of a string in a horizontal circle of 0.6-m radius so that the string make

s a 39 degree angle relative to the horizontal?
Physics
1 answer:
serious [3.7K]2 years ago
8 0

Answer:

2.7 m/s

Explanation:

Draw a free body diagram of the ball.  There are two forces:

Weight force mg pulling down

Tension force T pulling 39° above the horizontal

Sum of the forces in the y direction:

∑F = ma

T sin θ − mg = 0

T = mg / sin θ

Sum of the forces in the radial (+x) direction:

∑F = ma

T cos θ = m v² / r

Substitute:

(mg / sin θ) cos θ = m v² / r

mg / tan θ = m v² / r

g / tan θ = v² / r

v = √(gr / tan θ)

Given that r = 0.6 m and θ = 39°:

v = √(9.8 m/s² × 0.6 m / tan 39°)

v ≈ 2.7 m/s

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The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

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V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between  energies. This is the work-energy theorem. So,[tex]W = U_b - U_a

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3

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4 0
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Answer:

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Answer:

Explanation:

Conclusion is simple you can just say that it is the value written in words form only.

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8 0
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Answer:

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