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Lorico [155]
3 years ago
5

A fly flaps its wings back and forth 25 times in one second. The period of wing flapping is _____.

Physics
1 answer:
dalvyx [7]3 years ago
3 0

Answer:

<h2>The period is 0.04 sec</h2>

Explanation:

This problem is based on oscillatory motion.

What is the period of oscillation?

This can be defined as the time taken to complete one full revolution or oscillation.

given that the  fly flaps its wings back and forth 25 times in one second. then the frequency is 25 Hz

we know that period T= 1/F

T= 1/25

T=0.04 sec

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A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The
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Answer:

0.629\ \text{rad/s}^2 counterclockwise

9.98\ \text{s}

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\alpha_2 = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2

The angular acceleration of the pottery wheel is 0.629\ \text{rad/s}^2.

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

\omega_i = Initial angular velocity = 0

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t = Time taken

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}

The time it takes the pottery wheel to reach the required speed is 9.98\ \text{s}

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4 years ago
A 50 kg bobsled slides down an ice track starting (at zero initial speed) from the top of a(n) 184 m high hill. The acceleration
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Answer:

Explanation:

Given

mass of ice m=50\ kg

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