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3 years ago

. Draw the sketches of two waves A and B such that wave A has twice the wavelength and half the amplitude of wave B.

Physics

1 answer:

Pavel [41]3 years ago

4 0

Wave B will, therefore, be considered more powerful than wave A. Waves with higher frequencies are more energetic than those with lower frequencies. The equation below demonstrates this. The higher the amplitude also, the higher the energy of the wave.

v = f λ

Where,

v = velocity of the wave,

f = frequency of the wave,

λ = wavelength.

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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

aleksandrvk [35]

(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

Which explanation BEST supports Newton's second law of motion which states that force = mass x acceleration?

GaryK [48]

The answer is C. F=ma basically says that force is a function of mass multiplied by acceleration. The first two answers don’t make sense because there’s no necessary relationship between mass and acceleration. And for the last two, the higher the mass, the higher the force needed, therefore C is the correct answer.

8 0

3 years ago

A truck is moving around a circular curve at a uniform velocity of 13 m/s. If the centripetal force on the truck is 3,300 N and

Paladinen [302]

Answer: Option B.

Since here the truck is moving on a circular track, it will experience centripetal force.

F(centripetal) = m × acc
or
$r = \frac{m v^{2}}{F}$

where r is the radius of the track.
m is the mass of truck
v is the speed of the truck.
Given: v = <span>13 m/s
m = </span><span>1,600 kg
</span>F = 3300 Newton

To find = radius of track=?
$r = \frac{m v^{2} }{F}$
$r = \frac{1600*13*13}{3300}$
r = 81.94 m
Therefore, radius of track is 81.94 m

4 0

3 years ago

Read 2 more answers

The diagram below shows a wave with its wavelength indicated in red.

lisabon 2012 [21]

C. It will decrease

5 0

4 years ago

Read 2 more answers

The table shows data for four planetary bodies. If your mass is 68.05 kg, how

scZoUnD [109]

Answer: