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2 years ago

A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. what is the frequency of the wheel's rotation?

2 answers:

muminat2 years ago

8 0

The frequency is defined as the number of cycles done in one second:
 $f= \frac{N}{t}$ 
where N is the number of cycles and t is the time taken to complete N cycles. The wheel in this problem does N=2 revolutions in a time of t=4.0s, therefore its frequency of rotation is:
 $f= \frac{N}{t} = \frac{2.0}{4.0s} = 0.5 Hz$

Alex17521 [72]2 years ago

8 0

Answer: The frequency of the wheel's rotation is 0.5 Hz.

Explanation:

Frequency of rotation: It is defined as the number of revolution in a second.

The expression for the frequency in terms of the number of revolution in a given time is as follows;

$f=\frac{N}{t}$

Here N is the number of cycles and t is the time taken.

Put N= 2 rev and t= 4 s in the above expression.

$f=\frac{2}{4}$

$f=0.5 Hz$

Therefore, the correct answer is 0.5 Hz.

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To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

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The vertical component of velocity is

$-h = v_y t -\frac{1}{2} gt^2$

Here,

h = Height

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$-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2$

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$v_y = 21.77m/s$

The direction of the velocity will be given by the tangent of the components, then

$tan\theta = \frac{v_y}{v_x}$

$\theta = tan^{-1} (\frac{21.77}{14.89})$

$\theta = 55.59\°$

The magnitude is given vectorially as,

$|V| = \sqrt{v_x^2+v_y^2}$

$|V| = \sqrt{14.89^2 +21.77^2}$

$|V| = 26.37m/s$

Therefore the angle is 55.59° and the velocity is 26.37m/s

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