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Inessa05 [86]
3 years ago
12

II NEED HELP WITH THIS QUESTION PLS

Physics
1 answer:
Basile [38]3 years ago
4 0

Answer:

Suppose that the frequency of the first harmonic is F, then the frequency of the second harmonic will be 2*F

The third one will be 3*F

and so on.

Then:

a) the first harmonic is 262 Hz,

Then the third harmonic will be 3*262Hz = 786 Hz

b) if F is the frequency of the first harmonic, and we want to find this.

We know that the fifth harmonic frequency is 1700Hz

Then:

5*F = 1700 Hz

F = (1700Hz)/5 = 340Hz

c) We want to find the fifth harmonic such that we know that the third overtone frequency is 984 Hz.

An overtone is any frequency larger than the fundamental frequency; overtone is just another name for harmonic.

Such that the second overtone is the same as the second harmonic, the third overtone is the same as the third harmonic, etc.

Then the third overtone is the same as the third harmonic.

This means that if F is the first harmonic, then:

3*F = 984 Hz

F = 984Hz/3 = 328 Hz

And the fifth harmonic will be:

5*328 Hz = 1640 Hz

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Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

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b) velocity = 2m/s

time = 3.5s

Distance = 2m/s*3.5s = 7m

c) velocity = 10m/s

time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

e) velocity = 1.5m/s

time = 5s

Distance = 1.5m/s*5s = 7.5m

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3 years ago
A rocket starts from rest and moves upward from the surface of the earth. For the first 10s of its motion, the vertical accelera
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Since the rocket’s acceleration is 3.00 m/s^3 * t, its acceleration is increasing at the rate of 3 m/s^3 each second. The equation for its velocity at a specific time is the integral of the acceleration equation. 

<span>vf = vi + 1.5 * t^2, vi = 0 </span>
<span>vf = 1.5 * 10^2 = 150 m/s </span>
This is the rocket’s velocity at 10 seconds. The equation for its height at specific time is the integral velocity equation

<span>yf = yi + 0.5 * t^3, yi = 0 </span>
<span>yf = 0.5 * 10^3 = 500 meters </span>
<span>This is the rocket’s height at 10 seconds. </span>

<span>Part B </span>
<span>What is the speed of the rocket when it is 345 m above the surface of the earth? </span>
<span>Express your answer with the appropriate units. </span>


<span>Use the equation above to determine the time. </span>

<span>345 = 0.5 * t^3 </span>
<span>t^3 = 690 </span>
<span>t = 690^⅓ </span>
<span>This is approximately 8.837 seconds. Use the following equation to determine the velocity at this time. </span>

<span>v = 1.5 * t^2 = 1.5 * (690^⅓)^2 </span>
<span>This is approximately 117 m/s. </span>


<span>The graph of height versus time is the graph of a cubic function. The graph of velocity is a parabola. The graph of acceleration versus time is line. The slope of the line is the coefficient of t. This is a very different type of problem. For the acceleration to increase, the force must be increasing. To see what this feels like slowly push the accelerator pedal of a car to the floor. Just don’t do this so long that your car is speeding!!</span>
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In a nuclear power plant, Choose one: A. control rods regulate the rate of the reaction by releasing excess neutrons. B. pitchbl
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Answer:

option A.

Explanation:

The correct answer is option A.

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Control rods are inserted in the reactor if energy requirement is less and when energy requirement is large Control rods are remove. Hence, we can say that control rod can be used to regulate the reaction in the nuclear power plant.

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Answer:

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Initial speed of the car, u = 26.4 m/s

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To find,

The speed of the car after traveling this distance.

Solution,

The force experienced by a car is equal to the product of mass and acceleration.

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