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Pie
3 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]3 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]3 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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Answer:

160 Nm

Explanation:

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3 years ago
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Igoryamba

D. Free fall

Explanation:

An object is said to be in free fall when there is only one force acting on the body, which is the force of gravity.

Near the Earth's surface, the force of gravity acting on a body is given by

F = mg

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m is the mass of the body

g is the acceleration of gravity (its value is 9.8 m/s^2)

The direction of this force is downward (towards the Earth's centre).

If we apply Newton's second law on an object in free-fall, we can find its acceleration. In fact, we have:

a=\frac{F}{m}

And substituting F,

a=\frac{mg}{m}=g=9.8 m/s^2

So, every object in free-fall accelerates at 9.8 m/s^2 towards the ground.

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3 years ago
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Answer:

B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.

Explanation:

B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.

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2 years ago
irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano
Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

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Here for Regulus, we get:

S = 1 / p

  = 1 / (0.042)                                     [ 1 parsecs = 1 arcseconds ]

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We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

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