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lubasha [3.4K]
3 years ago
7

Two loudspeakers are about 10 mm apart in the front of a large classroom. If either speaker plays a pure tone at a single freque

ncy of 400 HzHz, the loudness seems pretty even as you wander around the room, and gradually decreases in volume as you move farther from the speaker. If both speakers then play the same tone together, what do you hear as you wander around the room
Physics
1 answer:
Yuliya22 [10]3 years ago
6 0

Answer:

I hear points of low volume sound and points of high volume of sound.

Explanation:

This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.

Since their frequencies are similar, we should have beats of high and low frequency.

So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.

Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.

So, as you wander around the room, I should hear points of high and low sound across the room.

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A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease
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The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

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

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm

So, the new pressure is 2 atm.

7 0
2 years ago
What distance will a car traveling at a speed of 15 m/s cover in 300 s?
AnnZ [28]

Distance formula: d = st

d = 15 * 300

d = 4,500m

Best of Luck!

3 0
3 years ago
A 4.0-kg object has 72 J of kinetic energy.<br> Determine its speed.
Svetlanka [38]

Answer:

6m/s

Explanation:

v = sqrt of 2KE/m

Where:

KE = kinetic energy

m = mass of a body

v = velocity of a body

= sqrt of 2(72)/4

= sqrt of 144/4

= sqrt of 36

= 6m/s

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2 years ago
In which of these cases is a waiter doing work on the object
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Section 2 is right,, i think. good luck
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2 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
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