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Vlad1618 [11]
3 years ago
8

A car enters a tunnel at 24 m/s and accelerates steadily at 2.0 m/s2. At what speed does it leave the tunnel, 8.0 seconds later?

Physics
1 answer:
Bogdan [553]3 years ago
4 0
Given: 

V1 = initial velocity = 24 m/s
a = acceleration = 2.0 m/s^2
s = time = 8 s
V2 = final velocity = ?

For linear-motion problems with those given terms, the following formula is used:

V2 = V1 + as

Substituting the given values:

V2 = 24 + 2(8)
V2 = 24 + 16
V2 = 40 m/s

Therefore the car will have a speed of 40 m/s as it leaves the tunnel.
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A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
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(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
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