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stealth61 [152]
3 years ago
10

What is the frequency of a wave when it speeds up at 5 m/s for a distance of 25 m?​

Physics
1 answer:
Arturiano [62]3 years ago
5 0

Answer:

<h3>0.2Hz= frequency </h3>

Explanation:

given=

v = 5m/s

distance = 25m

time= distance/ speed = 25/5

T= 5

and frequency = 1/T

= 1/5 = 0.2 Hz

hope this will help you

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b) The time taken by the disk to reach the final angular speed is 8 seconds.

Explanation:

a) Let suppose that the disk accelerates uniformly, given that initial and final angular speed (\omega_{o}, \omega_{f}), in radians per second, and change in angular position (\Delta \theta), in radians, are known. The angular acceleration (\alpha), in radians per square second, are found by using this expression:

\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta} (1)

If we know that \omega_{o} = 0\,\frac{rad}{s}, \omega_{f} = 15\,\frac{rad}{s} and \Delta \theta = 60\,rad, then the angular acceleration of the disk is:

\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}

\alpha = 1.875\,\frac{rad}{s^{2}}

The angular acceleration is 1.875 radians per square second.

b) The time taken by the disk to reach the final angular velocity is determined by the following kinematic formula:

t = \frac{\omega_{f}-\omega_{o}}{\alpha} (2)

Where t is the time, in seconds.

If we know that \omega_{o} = 0\,\frac{rad}{s}, \omega_{f} = 15\,\frac{rad}{s} and \alpha = 1.875\,\frac{rad}{s^{2}}, then the time taken by the disk is:

t = \frac{\omega_{f}-\omega_{o}}{\alpha}

t = 8\,s

The time taken by the disk to reach the final angular speed is 8 seconds.

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