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Aloiza [94]
3 years ago
15

Two cars, one in front of the other, are traveling down the highway at 53.7 m/s. The car behind sounds its horn, which has a fre

quency of 536 Hz. What is the frequency heard by the driver of the lead car
Physics
1 answer:
Vadim26 [7]3 years ago
4 0

Answer:

Explanation:

The source is approaching the observer which is moving away from the source , so applying Doppler's effect

n = n_o\frac{V-V_O}{V-V_S}

where n is apparent frequency , n₀ is original frequency , V₀ is velocity of observer , V_S is velocity of source , V is velocity of sound

Putting the given values

= n = 536\frac{340-53.7}{340-53.7 }

= 536

So apparent frequency will be same as original frequency .

ie 536 .

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A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m
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To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

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S_{el} = 414Mpa

Yield Strain of the Specimen

\epsilon_{el} = 0.002

Diameter of the test-specimen

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Gage length of the Specimen

L_0 = 50mm

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E = \frac{S_{el}}{\epsilon_{el}}

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E = 207Gpa

Strain energy per unit volume at the elastic limit is

U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}

U'_{el} = \frac{1}{2} (414)(0.002)

U'_{el} = 414kN\cdot m/m^3

Considering that the net strain energy of the sample is

U_{el} = U_{el}' \cdot (\text{Volume of sample})

U_{el} =  U_{el}'(\frac{\pi d_0^2}{4})(L_0)

U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})

U_{el} = 2.663N\cdot m

Therefore the net strain energy of the sample is 2.663N\codt m

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