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omeli [17]
3 years ago
6

A police car is parked by the side of a highway and its siren emits a sound measuring 750 Hz. As you approach the police car fro

m a distance, you hear a change in the sound. The change is in the sound's
Physics
2 answers:
Nikitich [7]3 years ago
8 0

As you approach the police car with blaring siren, you hear a change in pitch and frequency. The apparent pitch changes as a result of the Doppler effect. Frequency determines pitch; frequency and wavelength are inversely proportional. You wouldn't "hear" either one. The result is pitch.

D) pitch and frequency.

tangare [24]3 years ago
5 0

Answer:

The correct answer would be frequency, intensity, and amplitude.

It can be explained with the help of Doppler effects which states that the frequency or wavelength of sound changes (increases or decreases) as source and observer move towards or away from each other.

Due to this effect, the frequency and intensity of siren increase as we move towards the siren.

In addition, amplitude also increases as we move towards the source of the sound and decreases as the observer moves away from the source.

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The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
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Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

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Answer:

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