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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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Speed=distance/time

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speed=48/2
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speed=24km/h

FORMULAS

speed = distance/time
time = distance/speed
distance = speed×time

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Explanation:

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as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

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Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

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At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

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