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natta225 [31]
3 years ago
5

Suppose you were bungee jumping from a bridge while blowing a hand-held air horn. How would someone remaining on the bridge hear

the pitch of the air horn as time increased?
Physics
2 answers:
castortr0y [4]3 years ago
7 0

Answer:

Pitch would decrease

Explanation:

According to Doppler's effect, the apparent frequency (pitch) of a sound changes if there is relative motion between the source and the listener.

As the distance increases between the source and the listener, the apparent frequency heard by the listener would be lower than the source frequency.

So, as the bungee jumper would go up in the air, the listener standing on the bridge would hear decreasing pitch.

Genrish500 [490]3 years ago
4 0

Answer:

The pitch will progressively lower

Explanation:

If i were bungee jumping from  a bridge while blowing a hand-held air horn and someone who remains on the bridge will hear a decreased pitch or frequency as the source is moving away from the stationary listener as per the Dooplers effect. Hence, the pitch will progressively lower as the source is moving away from the observer.

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Which one of the following temperatures is equal to 5°C?
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Answer : The correct option is, (D) 278 K

Explanation :

We are given temperature 5^oC.

Now the conversion factor used for the temperature is,

K=^oC+273

where, K is kelvin and ^oC is Celsius.

Now put the value of temperature, we get

K=5^oC+273=278K

Therefore, the temperature 278 K is equal to the 5^oC


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A baseball goes from zero to 38 m/s in 0.157 s. what is its average acceleration? answer in units of m/s 2 .
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38m/s x 6.369 = 242.038 m/s^2
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this is a link to a web sight with a diagram to help you

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What will happen if you didn’t have chemical energy in your body
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If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


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8 0
3 years ago
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