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seraphim [82]
3 years ago
8

I need some help w/ this:

Physics
1 answer:
pochemuha3 years ago
5 0

Answer:

The velocity of the police car is, v = 17.798 m/s

Explanation:

Given data,

The actual frequency of the siren, f = 2010 Hz

The observed frequency of siren is, f' = 2120 Hz

The velocity of the observer, v' = 0 m/s

The velocity of the source, v = ?

The formula for Doppler effect,

                            f'=\frac{(V+v')}{(V-v)}f

Where,

                         V - velocity of sound waves in air.

                          v=V-(V+v')\frac{f}{f'}

Substituting the given values,

                         v=343-(343+0)\frac{2010}{2120}

                                 v = 17.798 m/s

Hence, the velocity of the police car is, v = 17.798 m/s

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This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

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a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

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b) the tangential acceleration of the ball is;

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Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

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we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

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