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DanielleElmas [232]
1 year 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]1 year 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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Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
2 years ago
160 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabol
anastassius [24]

Answer:

minimum power should be used to operate the air conditioner is 4000 W

Explanation:

given data

students  n = 160

power p = 125 W

COP = 5.0

to find out

what minimum power should be used

solution

we know the COP formula that is given below

COP = students × power  / minimum power

minimum power = n × p / COP

put all value

minimum power = n × p / COP

minimum power = 160 × 125 / 5

minimum power = 4000 W

minimum power should be used to operate the air conditioner is 4000 W

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A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

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V = Vo + a.t&#10;&#10;

The ball is against the vector of gravity. Then, the gravity will be negative.

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The ball will stop in the air after approx. 4.72 seconds. And will take the same time to hit the ground.

It will stay approx. 9.44 seconds in the air.
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GarryVolchara [31]
Answer: Δp = F*Δt = 500N*0.001s = 0.5Ns
3 0
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