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DanielleElmas [232]
2 years ago
9

A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a

car is moving away from this stereo at a speed of 40.0 m/s. The frequency of sound waves that the car receives is ________. In addition, when the car is 70.0 meters away from the speaker, the car will hear sound waves with a sound intensity level of _________ .
Physics
1 answer:
photoshop1234 [79]2 years ago
3 0

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

f'=f(\frac{v-v_o}{v+v_s})    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

v=\lambda f\\\\

v: speed of sound

λ: wavelength = 0.700 m

f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz

Next, you replace the values of all parameters in the equation (1):

f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

I=\frac{P}{4\pi r^2}

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}

hence, the intensity is 8.12*10^⁻4 W/m^2

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A small rivet connecting two pieces of sheet metal is being clinched by hammering. Determine the impulse exerted on the rivet an
kykrilka [37]

Answer:

a) the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

the energy absorbed by the rivet under each blow  when the anvil has an infinite mass support = 9.32 ft.lb

b) the impulse exerted by the rivet when the anvil has a support weight of 9 lb = 0.799 lb.s

the energy absorbed by the rivet under each blow when the anvil has a support weight of 9 lb is = 7.99 ft.lb

Explanation:

The first picture shows a schematic view of a free body momentum diagram of the hammer head and the anvil.

Using the principle of conservation of momentum to determine the final velocity of anvil and hammer after the impact; we have:

m_Hv_H + m_Av_A = m_Hv_2+m_Av_2

From the question given, we can deduce that the anvil is at rest;

∴ v_A = 0; then, we have:

m_Hv_H + 0 = (m_H+m_A) v_2

Making v_2 the subject of the formula; we have:

v_2 = \frac{m_Hv_H}{m_H + m_A}       ------- Equation  (1)

Also, from the second diagram; there is a representation of a free  body momentum  of the hammer head;

From the diagram;

F = impulsive force exerted on the  rivet

Δt = the change in time of application of the impulsive force

Using the principle of impulse of momentum to the hammer in the quest to determine the impulse exerted (i.e FΔt ) on the rivet; we have:

m_Hv_H - F \delta t = m_Hv_2

- F \delta t = - m_Hv_H + m_Hv_2

F \delta t = m_Hv_H - m_Hv_2

F \delta t = m_H(v_H - v_2)        ------- Equation   (2)

Using the function of the kinetic energy  of the hammer before impact T_1; we have:

T_1 = \frac{1}{2} m_Hv_H^2  -------- Equation (3)

We determine the mass of the hammer m_H  by using the formula from:

W_H = m_Hg

where;

W_H = weight of the hammer

m_H = mass of the hammer

g = acceleration due to gravity

Making m_H the subject of the formula; we have:

m_H = \frac{W_H}{g}

m_H = \frac{1.5 \ lb}{32.2 \ ft/s^2}

m_H = 0.04658 \ lb.s^2/ft

Now;

T_1 = \frac{1}{2} m_Hv_H^2

T_1 = \frac{1}{2}*(0.04658 \ lb.s^2 /ft) *(20 \ ft/s)^2

T_1 = \frac{18.632 }{2}

T_1 = 9.316 \ ft.lb

After the impact T_2 ; the final kinetic energy of the hammer and anvil can be written as:

T_2 = \frac{1}{2}(m_H +m_A)v^2_2

Recall from equation (1) ; where v_2 = (\frac{m_Hv_H}{m_H+m_A})  ; if we slot that into the above equation; we have:

T_2 = \frac{1}{2}(m_H +m_A)( \frac{m_Hv_H}{m_H+m_A})^2

T_2 = \frac{1}{2} \frac{m^2_H +v^2}{m_H+m_A}

T_2 = \frac{1}{2} ({m^2_H +v^2})(\frac{m_H}{m_H+m_A})

Also; from equation (3)

T_1 = \frac{1}{2} m_Hv_H^2; Therefore;

T_2 = T_1 (\frac{m_H}{m_H+m_A})    ----- Equation (4)

a)

Now; To calculate the impulse exerted by the rivet FΔt and the energy absorbed by the rivet under each blow  ΔT when the anvil has an infinite mass support; we have the following process

First , we need to find the mass of the anvil when we have an infinite mass support;

mass of the anvil m_A = \frac{W_A}{g}

where we replace;  W_A \ with \ \infty and g = 32.2 ft/s²

m_A =  \frac{\infty}{32.2 \ ft/s}

However ; from equation (1)

v_2 = \frac{m_H v_H}{m_H + m_A}

v_2 = \frac{0.04658*20}{0.04658+ \ \infty}

v_2 = 0

From equation (2)

F \delta t = m_H(v_H + v_2)      

F \delta t = (0.04658 lb .s^2 /ft )(20ft/s  - 0)

F \delta t = \ 0.932 \  lb.s

Therefore the impulse exerted by the rivet when the anvil has an infinite mass support is  0.932 lb.s

For the energy absorbed by the rivet ; we have:

T_2 = T_1 (\frac{m_H}{m_H+m_A} )

where;

T_1= 9.316 \ ft.lb

m_H = 0.04658 \ lb.s^2/ft

m_A = \infty

Then;

T_2 = (9.316 \ ft.lb) (\frac{0.04658\  lb.s^2/ft)}{0.04658  \ lb.s^2/ft+ \infty} )

T_2 = (9.316 \ ft.lb)* 0

T_2 = 0

Then the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support

ΔT = T_1 - T_2

ΔT = 9.316 ft.lb - 0

ΔT ≅  9.32 ft.lb

Therefore; we conclude that the energy absorbed by the rivet under each blow  when the anvil has an infinite mass support = 9.32 ft.lb

b)

Due to the broadness of this question, the text is more than 5000 characters, so i was unable to submit it after typing it . In the bid to curb that ; i create a document for the answer  for the part b of this question.

The attached file can be found below.

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We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

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