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Effectus [21]
4 years ago
10

If a fast-moving car making a loud noise approaches and moves past the person, what will happen as the distance between the two

increases?
A. The pitch of the sound being heard by the person will appear to be higher than the pitch of the source.
B. The frequency of the sound waves reaching the person's ear will be greater than the frequency of the waves leaving the car.
C. The pitch of the sound being heard by the person will appear to be lower than the pitch of the source.
D. The pitch and frequency of the sound waves reaching the person's ear will remain unchanged.
Physics
2 answers:
densk [106]4 years ago
5 0
C.  The Pitch of the sound being heard by the person will appear to be lower than the pitch of the source.  I hope this is right

vladimir2022 [97]4 years ago
3 0
The answer is B.
The frequency will be greater


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3 years ago
The velocity of an object includes its speed and what
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I believe that the correct answer you are looking for is the distance traveled 

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3 years ago
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A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t
andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = \frac{r}{2}

f = \frac{0.40}{2} = 0.20 m

Now,

m = - \frac{v}{u}

- 2 = -\frac{v}{u}

\frac{v}{u} = 2                  (2)

Now, by lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

v = \frac{uf}{u - f}            (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = \frac{uf}{u - f}

2 = \frac{f}{u - f}

2(u - 0.20) = 0.20

u = 0.30 m

6 0
3 years ago
A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t
fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

A \rightarrow (A-\frac{A}{9} )

A \rightarrow \frac{8A}{9}

As we know,

Energy will be:

⇒  E_{1}=\frac{1}{2}KA^2

and,

⇒  E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2

          =\frac{64KA^2}{162}

⇒  \Delta E=E_1-E_2

On putting the estimated values, we get

           =\frac{1}{2}KA^2-\frac{64KA^2}{162}

⇒  \frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}

          =\frac{40}{162}

          =0.246

3 0
3 years ago
Convert 1erg into joule by dimensional method​
Valentin [98]

Answer:

1 * 10^-7 [J]

Explanation:

To solve this problem we must use dimensional analysis.

1 ergos [erg] is equal to 1 * 10^-7 Joules [J]

1[erg]*\frac{1*10^{-7} }{1}*[\frac{J}{erg} ] \\= 1*10^{-7}[J]

4 0
3 years ago
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