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xeze [42]
3 years ago
9

To practice Problem-Solving Strategy 16.2 Doppler effect. The sound source of a ship’s sonar system operates at a frequency of 2

2.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water. mastering physics g
Physics
1 answer:
Whitepunk [10]3 years ago
7 0

Answer:

147 Hz

Explanation:

First we need to know what frequency will the waves hitting the whale have respect of it.

The change in frequency is calculated with this equation:

\Delta f = \frac{\Delta v}{c} * f0

However in this case, the wave is bouncing on the whale and coming back, so there is a double frequency shift (the difference is the same both times because the relative speeds between the two objects is the same both times)

\Delta f = 2*\frac{\Delta v}{c} * f0

Then:

\Delta f = 2*\frac{\4.95 - 0}{1482} * 22000 = 147 Hz

The frequency shift is positive, it will have increased frequency.

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Hey there!

The answer is : 
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4 0
2 years ago
A meteor moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement describ
Thepotemich [5.8K]

Answer:

this statement describes meteor's velocity,

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3 0
3 years ago
A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet
kogti [31]
The centripetal force, Fc, is calculated through the equation, 
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4 0
3 years ago
Read 2 more answers
A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee
rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water =  700-kg

length of cable = 20 m

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angle of the cable = 38.0°

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T sin 38^0= \dfrac{mv^2}{l} + F..........(2)

equation (1)/(2)

tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}

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           F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0
3 years ago
How many joules of work are done against a truck when a force of 50 N pushes it 1 kilometer away
ELEN [110]

Answer:

Work = 50,000 J

Explanation:

Work = force * distance

Given that,

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Work = 50 * 1000

Work = 50,000 J

8 0
2 years ago
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