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ICE Princess25 [194]
3 years ago
6

(4%) Problem 10: The standard frequency of the musical pitch called "middle C" is 261.6 Hz. show answer No Attempt How fast, in

meters per second, would a source emitting a tone of middle C have to move toward you so that you would hear the pitch C# (C sharp) above middle C, with a frequency of 277.2 Hz? Assume that takes place in air at a high altitude, where the speed of sound is 298 m/s.
Physics
1 answer:
bearhunter [10]3 years ago
6 0

Answer:

16.77 m/s

Explanation:

Given that

Frequency of middle pitch, Fo = 261.6 Hz

Frequency of C sharp, f = 277.2 Hz

Velocity of sound in air, v = 298 m/s

Speed of sound from the source, Vs = ? m/s

Using the formula

f = Fo•(V + Vr)/(V + Vs)

← Doppler

Vr would be +ve if the receiver is moving toward source;

Vs would be -ve if source is moving toward the receiver

277.2 Hz = 261.6Hz * (298 + 0) / (298 - Vs)

277.2 = 77956.8 / (298 - Vs)

298 - Vs = 77956.8 / 277.2

298 - Vs = 281.23

Vs = 298 - 281.23

Vs = 16.77 m/s

Thus, the speed needed is 16.77 m/s

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Answer: True


hope this helps!

4 0
2 years ago
A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with
NNADVOKAT [17]

Answer: 1.59 sec

Explanation:

The kinetic energy contained in the rotation of the cylinder is 1/2 m v^2

The kinetic energy of translation is also 1/2 m v^2

so the total energy is m v^2

The force applied is mg sin (theta)

= m x 9.8 x 1/2

= 4.9 m

Now equate

F x d = m v^2

4.9 m x 3.1 = m v^2

v^2 = 4.9 x 3.1

v = sqrt(4.9 x 3.1) = 3.9 m/s

Acceleration.

V^2 = 2 a s

4.9 x 3.1 = 2 x a x 3.1

4.9 = 2 a

a = 2.45 m/s^2

Time

T = v/a

= 3.9/2.45

= 1.59 sec

6 0
4 years ago
A 60kg block is at rest on a ramp inclined at 35 degrees.
weeeeeb [17]

Answer:

Explanation:

(a)Gravitational Force is W=mg

Frictional force is f

Normal force is N

(b)Since the block is in Equilibrium therefore from diagram we can see that  

f=mg\sin \theta

f=60\times 9.8\times \sin 35

f=337.26 N

(c)Magnitude of Normal Force

N=mg\cos \theta =481.66 N

(d)friction is also given by

f=\mu mg\cos \theta

and f=mg\sin \theta

thus tan \theta =\mu

\mu =0.7  

8 0
3 years ago
For what visible wavelengths of light do the reflected waves interfere constructively? The range of wavelength of visible light
Otrada [13]

Answer:

None of the wavelength is in the visible range

Explanation:

Constructive interference of the reflected waves for different wavelengths can be estimated using:

λ_{m} = 2nd/m

where m is 1,2,3, ...

Therefore:

m=1, λ_{1} = 750 nm

m=2, λ_{1} = 750/2 = 375 nm

The limits of eye's sensitivity is between 430 nm and 690 nm. Beyond this range, the eye's sensitivity drops to approximately 1% of its maximum value.

7 0
4 years ago
Directions: I answer all the questions listed below.
anzhelika [568]

Answer:

The compression ratio is 10

Explanation:

Given

V_d = 450cm^3 --- swept volume

V_c = 50cm^3 --- compression volume

Required

Determine the compression ratio (CR)

This is calculated as:

CR = \frac{V_d + V_c}{V_c}

CR = \frac{450+50}{50}

CR = \frac{500}{50}

CR = 10

5 0
3 years ago
Read 2 more answers
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