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earnstyle [38]
3 years ago
8

What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from

its siren
Physics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer:

77.6 Hz

Explanation:

What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren.

The given parameters are:

Observer speed = 115 km/h

Source frequency = 753 Hz

Speed of sound = 342 m/s

First convert km/h to m/s

Observer speed = (115 × 1000) / 3600

Observer speed = 31.94 m/s

The frequency received by the person will be:

F = fv / ( V - v )

Where

F = frequency received by the person

f = siren frequency

V = speed of sound

v = speed of the ambulance

Substitute all the parameters into the formula

F = (753 × 31.94) / ( 342 - 31.94 )

F = 24050.82 / 310.06

F = 77.568 Hz

Therefore, the frequency received by a person is approximately 77.6Hz

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Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

       L₀ = I₁ w₁

final moment. After releasing the bricks

       L_{f} = I₂W₂

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       I₁ w₁ = I₂ w₂

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     w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

 

 let's calculate

       w₂ = 6.0/2.0   7.54

       w₂ = 22.6 rad/s

3 0
3 years ago
Where are the nucleus and the electrons located in this atom
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The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr
stealth61 [152]

Answer:

\frac{R_2}{R_1}=\frac{A_1}{A_2}\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

Explanation:

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R=\frac{\rho L}{A}

\rho is the material's resistance, L is the legth and A is the cross-sectional area.

For the first and second coils, we have:

R_1=\frac{\rho L}{A_1}\\R_2=\frac{\rho L}{A_2}\\\rho L=R_1A_1\\\rho L=R_2A_2\\R_1A_1=R_2A_2\\\frac{R_2}{R_1}=\frac{A_1}{A_2}

For the third and fourth coils, we have:

R_3=\frac{\rho L'}{A_3}\\R_4=\frac{\rho L'}{A_4}\\\rho L'=R_3A_3\\\rho L'=R_4A_4\\R_3A_3=R_4A_4\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

6 0
3 years ago
While watching a recent science fiction movie, one Klingon spaceship blows up a Droid spaceship with a laser gun. The Klingon cr
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The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor
uysha [10]

Answer:

c.a_m

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We are given that

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Acceleration due to gravity on the earth=a_e

g_m=\frac{1}{6}g_e

Net force due to am on an object on moon=F_{net}=ma_m

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a_e=a_m

Hence, option c is true.

3 0
3 years ago
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