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ivanzaharov
2 months ago
12

A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.

The string makes a constant angle of 61.0 ∘ with the vertical as the ball moves at a constant speed in a horizontal circle.If it takes the ball 1.25 s to complete one revolution, what is the magnitude of the radial acceleration of the ball?
Physics
1 answer:
Sav [3.1K]2 months ago
7 0

Answer:

a = 17.68 m/s²

Explanation:

Given:

Length of the string, L = 0.8 m

Angle with vertical, θ = 61°

Time for one complete revolution, t = 1.25 s

Required: radial acceleration =?

First, we must calculate the radius of the circular motion.

R = L sin θ

R = 0.8 x sin 61°

R = 0.7 m

Next, we compute the angular velocity.

\omega =\dfrac{2\pi}{T}

\omega =\dfrac{2\pi}{1.25}

ω = 5.026 rad/s

Now calculating the radial acceleration:

a = r ω²

a = 0.7 x 5.026²

a = 17.68 m/s²

Therefore, the radial acceleration of the ball is 17.68 m/s²

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It is known that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v are constant on both sides. Therefore we have

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let the final pressure be P and the temperature T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}..................2

similarly

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divide equation (2) by equation (3)

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T_1 {f} = 1.48 T_2 {f}

thus, the left side temperature equals 1.48 times the right side temperature

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