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4 years ago

2. A certain object revolves at a rate of 30 rpm. Please determine the frequency and period of this

Physics

1 answer:

schepotkina [342]4 years ago

7 0

Answer:

The time period of the motion is, T = 0.03 s

The frequency of the rotation is, f = 30 Hz

Explanation:

Given data,

The rotational speed of an object, ω = 30 rpm

ω = 188.5 rad/s

The time period of motion is,

T = 2π / ω

Substituting the given values in the above equation

= 2π / 188.5

T = 0.03 s

The time period of the motion is, T = 0.03 s

The frequency of rotation,

f = 1 /T

= 1 / 0.03

= 30 Hz

Hence, the frequency of the rotation is, f = 30 Hz

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3 years ago

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

frozen [14]

Answer:

$\theta = 49.81^0$

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Given that:

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$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

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3 years ago

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3 years ago

Jayne lifts the barbell 120 cm upwards. She has a mass of 60kg. How much work does she do?

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4 years ago

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kramer

Answer:

Explanation:

Given:

- Acceleration a_dry = 0.4*a_wet

- initial velocity V_i

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By how much does time taken to stop of dry snow differ from time taken to stop on wet snow?

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t_wet = - a_wet / V_i

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t_dry = - a_dry / V_i

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t_dry - t_wet = - a_dry / V_i + a_wet / V_i

dt = (a_wet - a_dry) / V_i

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3 years ago