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Arturiano [62]
2 years ago
5

Which situations might cause two observers (A and B) to measure different frequencies for the same vibrating object? Select the

two correct answers. (2 points)
Observer A and Observer B are both stationary and at the same distance from the object.
Observer A and Observer B are both stationary and at the same distance from the object.

Observer A is stationary and Observer B is moving.
Observer A is stationary and Observer B is moving.

Observer A and Observer B are moving at different speeds relative to each other.
Observer A and Observer B are moving at different speeds relative to each other.

Observer A and Observer B are stationary but at different distances from the vibrating object.
Physics
1 answer:
Alex787 [66]2 years ago
6 0

We want to explain why two different observes may measure different frequencies for the same vibrating object.

We will see that the two correct options are:

  • <em>Observer A is stationary and Observer B is moving.</em>
  • <em>Observer A and Observer B are moving at different speeds relative to each other.</em>

<em />

Let's assume that the vibrating object is a guitar string. Thus, the string makes a noise, and from that noise, we can estimate the frequency at which the string vibrates.

Now there appears a really cool effect, called the Doppler Effect. It says that the apparent change of frequency is <u>due to the motion of the observer or the source of the frequency (or both).</u>

For example, if you move towards the vibrating string, the perceived frequency will be larger, and you will hear a "higher" sound.

While if you move away from the string, the opposite happens, and you will hear a "lower" sound.

Then the only thing that impacts in how we perceive the frequency is our velocity relative to the source.

So, why do observers A and B measure different frequencies?

The two correct answers are:

  • <em>Observer A is stationary and Observer B is moving.</em>
  • <em>Observer A and Observer B are moving at different speeds relative to each other.</em>

If you want to learn more, you can read:

brainly.com/question/17107808

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1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

Explanation:

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the electric potential energy of the electron when it is  at the midpoint

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F = k \frac{q_{e}q}{r}

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k = constant (8.99 x 10^{9} Nm^{2} /C^{2})

electron charge, q_{e} = - 1.6 x 10^{-19} C

since it is measured at the midpoint,

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F = F_{1}+ F_{2}

  = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

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the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

r_{1} = 10 cm = 0.1 m

r_{2} = 0.5 - 0.1 = 0.4 m

F = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = kq_{e}(\frac{q_{1} }{r_{1} }+\frac{q_{2} }{r_{2} })

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} /0.1+2 x 10^{-9}/0.4)

  = - 5.04 x  10^{-17} J

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