Physics, 12.03.2020 05:50 leo4687

A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certain frequency. You stand far from each merry-go-round. The frequency you hear for each of your friend's three rides varies as the merry-go-round rotates. The variations in frequency for the three rides are given by the three curves in Fig. 17.5. Rank the curves according to (a) the linear speed of the sound source, (b) the angular speeds of the merry-go-rounds, and (c) the radii r of the merry-go-rounds, greatest first.

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Answer from: lalisa6745

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The Ranking of the curve according to their speed would be equal Rank because v_1 =v_2 =v_3

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

w_1 w_2 = w_3

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

r_2 =r_3 r_1

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency F_1 = F

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because v_1 =v_2 =v_3

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

w_1 w_2 = w_3

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

v = wr

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

r_2 =r_3 r_1

A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that em

Answer from: Quest

It should be d if not then c

Answer from: Quest

5.5 n====== $\sum f = ma$ assuming that $\sum f = f_{app}$ we have m = 0.145 kg a = 38 m/s²therefore $f_{app} = (0.145\text{ kg})(38\text{ m/s}^2) = 5.51\text{ n}$ 5.51 n, but two significant figures.final answer: 5.5 n

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