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neonofarm [45]
3 years ago
6

Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, su

ch as sound and water waves. Suppose that your physics professor sets up two sound speakers in the front of your classroom and uses an electronic oscillator to produce sound waves of a single frequency. When she turns the oscillator on (take this to be its original setting), you and many students hear a loud tone while other students hear nothing.
The professor adjusts the oscillator to produce sound waves of twice the original frequency. What happens?

a. Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing.
b. The students who originally heard a loud tone again hear a loud tone, and the students who originally heard nothing still hear nothing.
c. Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone.
d. The students who originally heard a loud tone now hear nothing, and the students who originally heard nothing now hear a loud tone.

Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

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b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

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= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

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Total energy =

2Vf² - 0.024

Using conservation of energy,

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22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

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\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

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