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ycow [4]
3 years ago
13

Sound with frequency 1300 Hz leaves a room through a doorway with a width of 1.03 m . At what minimum angle relative to the cent

erline perpendicular to the doorway will someone outside the room hear no sound
Physics
1 answer:
AlexFokin [52]3 years ago
6 0

Answer:

  about 14.7°

Explanation:

The formula for the angle of the first minimum is ...

  sin(θ) = λ/a

where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.

The wavelength of the sound is the speed of sound divided by the frequency:

  λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m

Then the angle of interest is ...

  θ = arcsin(0.261538/1.03) ≈ 14.7°

At an angle of about 14.7°, someone outside the room will hear no sound.

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A car has an acceleration of -5 m/s^2. Describe the car’s motion
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Explain two scenarios where a large truck can have the same momentum as a small car.
KengaRu [80]

The momentum, p, of any object having mass m and the velocity v is

p=mv\cdots(i)

Let M_L and M_S be the masses of the large truck and the car respectively, and V_L and V_S be the velocities of the large truck and the car respectively.

So, by using equation (i),

the momentum of the large truck = M_LV_L

and the momentum of the small car = M_SV_S.

If the large truck has the same momentum as a small car, then the condition is

M_LV_L=M_SV_S\cdots(ii)

The equation (ii) can be rearranged as

\frac {M_L}{M_S}=\frac {V_S}{V_L} \; or \; \frac{M_L}{V_S}=\frac{M_S}{V_L}

So, the first scenario:

\frac {M_L}{M_S}=\frac {V_S}{V_L}

\Rhghtarrow M_L:M_S=V_S:V_L

So, to have the same momentum, the ratio of mass of truck to the mass of the car must be equal to the ratio of velocity of the car to the velocity of the truck.

The other scenario:

\frac{M_L}{V_S}=\frac{M_S}{V_L}

\Rhghtarrow M_L:V_S= M_S:V_L

So, to have the same momentum, the ratio of mass of truck to the velocity of the car must be equal to the ratio of mass of the car to the velocity of the truck.

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a 60 kg woman in a elevator is accelerating upward at a rate of 1.2 m/s2. What is the gravitational force acting upon the woman?
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The gravitational force acting upon the woman is equal to <u>-588.6N</u>

Why?

To solve the problem, we need to consider that two forces are acting upon the woman, the first one is related to her weight and the second one is related to the acceleration of the elevator.

Gravitational force acting upon the woman:

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Hence, we have that the gravitational force acting upon the woman is equal to -588.6N.

Have a nice day!

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