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Pie
4 years ago
5

Suppose these waves represent the sound of a siren on a passing ambulance. Which wave represents the sound of the siren after it

has passed you? Explain your answer.
The wave A has a higher frequency and wave B has a lower frequency.

I REALLY need help. Please.
Physics
2 answers:
sergeinik [125]4 years ago
4 0
The answer is wave B has a lower frequency.  Suppose these waves represent the sound of a siren on a passing ambulance. Wave B (lower frequency) represents the sound of the siren after it has passed you.  The ambulance moving towards the observer produces a higher than normal frequency, and the ambulance moving away produces a lower than normal frequency.
sergey [27]4 years ago
4 0

Answer:

The correct answer is "wave B has a lower frequency."

Explanation:

The Doppler effect is defined as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave movement when the sender and the receiver, or observer, move relative to each other.

This is the effect that occurs in this case. And it is explained as follows: When the ambulance approaches the observer, the wave "compresses", that is, the wavelength is short, the frequency high and, therefore, the tone of the perceived sound will be sharp. On the other hand, when the ambulance moves away from the observer, the wave "decompresses", that is, the wavelength is long and the frequency low.

Then the correct answer is "wave B has a lower frequency."

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What do all hydrogen atoms and ions have in common?
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3 years ago
Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo
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To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.

By definition we know that

KE + KR = PE

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh

Where,

m = mass

v= Velocity

I = Moment of Inertia

\omega = Angular velocity

g = Gravity

h = Height

We know as well that \omega = v/r for velocity (v) and Radius (r)

Therefore replacing we have

\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh

[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2[/tex]

h= \frac{1}{2}v^2 ( \frac{1}{g} +\frac{I}{mg}\frac{1}{r^2} )

h= \frac{1}{2}0.75^2 ( \frac{1}{9.8} +\frac{2.9*10^{-5}}{(0.056)(9.8)}\frac{1}{(0.0064)^2} )

h = 0.3915m

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6 0
3 years ago
5. A cable is attached 32.0 m from the base of a flagpole that is about to
soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, F_v = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

F_v + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

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R ≈ 30,382.373 N

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