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2 years ago

3

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 49.3° with the vertical as the ball moves at a constant speed in a horizontal circle. If it takes the ball 1.45 s to complete one revolution, what is the magnitude of the radial acceleration of the ball?

1 answer:

vitfil [10]2 years ago

3 0

Answer:

a= 11.36 m/s²

Explanation:

Radius of circle=R=0.8 sin∅

=0.8 sin(49.3)

=0.606 m

and we know that

ω= $\frac{2\pi }{T}$

= $\frac{2*3.141592}{1.45}$

ω=4.33 $s^{-1}$

a=R ω²

=0.606*( $4.33^{2}$

a=11.36 m/s²

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Answer:

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(b)

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(c)

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2 years ago

A brass lid screws tightly onto a glass jar at 20 degrees C. To help open the jar, it can be placed into a bath of hot water. Af

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Answer:

0.0016 cm

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$\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm$

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Sav [38]

Answer:

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AlexFokin [52]

Answer:

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