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Zarrin [17]
3 years ago
10

A sound wave has a frequency of 425 Hz. What is the period of this wave?

Physics
1 answer:
Serhud [2]3 years ago
8 0
Period is 1/frequency
1/425 = 2.353ms
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How does energy transform from one form to another?
Sliva [168]
Energy can only be converted from one form to another

eg kinetic energy is converted to heat energy
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3 years ago
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A resistor converts 350 J of electrical energy to other forms of energy. What is the amount of
Helga [31]

Answer:

50C

Explanation:

Given parameters:

Electrical energy  = 350J

Potential difference  = 7V

Unknown:

Amount of charge = ?

Solution:

To solve this problem, use the expression below;

  E  = q x v

E is the electrical energy

q is the quantity of charge

v is the voltage

Insert the parameters and solve for q;

            350 = q x 7

            q = \frac{350}{7}

             q = 50C

5 0
3 years ago
If the mass of an object is 100 and Net force is 400, what would be the Acceleration?
ira [324]

Answer:

4m/s^{2}

Explanation:

I'm assuming the units for force and mass are Newtons and kilograms, respectively.

Rearranging Newton's first law (F=m*a) to solve for acceleration:

F=m*a

a=\frac{F}{m}=\frac{400 N}{100 kg}=4m/s^{2}

The acceleration is 4 meters per second squared and was found by rearranging Newton's first law in order to solve for acceleration.

8 0
3 years ago
What is the nuclear binding energy in joules for an atom of helium–4? Assume the following: Mass defect = 5.0531 × 10-29 kilogra
Romashka [77]
The equation given, E = mc², can be directly used to determine the unknown amount of nuclear binding energy. Substitute the values given to the equation,
               E = (5.0531 x 10^-29 kg) x (3 x 10^8 m/s)² = 4.548 x 10^-12 J
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5 0
4 years ago
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Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz
Olin [163]

Answer:

The correct answer is option 'd': The frequency decreases and the intensity of the sound decreases.

Explanation:

1) <u>Effect on Frequency </u>

According to Doppler's effect of sound we have

for a source of sound moving away from the observer the relation between the observed and the original frequency is given by

f_{app}=\frac{c-v_{rec}}{c+v_{s}}\times f_{original}

where

c = speed of sound in air

v_{rec} is the velocity of observer of sound

v_{s} is the velocity of source of sound

f_{o} is the original frequency of sound

As we see the ratio is less than 1 thus the frequency of sound that the observer receives is less than that of source.

2) <u>Effect on Intensity:</u>

At a distance 'r' from source emitting a wave of Power 'P' is given by

I=\frac{P}{4\pi r^{2}}

As we see on increasing 'r' intensity of sound decreases.

3 0
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