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3 years ago

8. What is the frequency heard by a person driving at 18 m/s toward a blowing factory whistle (600 Hz) if

Physics

1 answer:

Angelina_Jolie [31]3 years ago

7 0

Answer:

631.7 Hz

Explanation:

The Doppler effect is a phenomenon that occurs when there is relative motion between an observer and a source of a wave. When this occurs, the is a change in the apparent frequency of the wave, as observed by the observer.

In particular, the new apparent frequency can be calculated as:

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the real frequency

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if the observer is moving towards the source, negative if moving away)

$v_s$ is the velocity of the source (negative if the source is moving towards the observer, positive if moving away)

In this problem:

v = 340.6 m/s is the speed of the sound wave

f = 600 Hz is the real frequency of the sound

$v_o = +18 m/s$ is the velocity of the person

$v_s=0$ is the velocity of the source

Substituting, we find:

$f'=\frac{340.6+18}{340.6}(600)=631.7 Hz$

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NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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