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Fittoniya [83]
3 years ago
9

A sound wave in the air has a frequency of 256 hertz and is traveling at a speed of 331 meters/second. What is its wavelength?

Physics
2 answers:
nikdorinn [45]3 years ago
6 0
2.39 is the answer cuh
emmainna [20.7K]3 years ago
6 0

Answer: The correct answer is 1.29 m/s.

Explanation:

The expression for the velocity of the wave is as follows;

v=f\lambda

Here, f is the frequency of the wave, v is the velocity of the wave and \lambda is the wavelength of the wave.

It is given in the problem that a sound wave in the air has a frequency of 256 hertz and is traveling at a speed of 331 meters/second.

Calculate the wavelength of the wave by rearranging the above expression.

Put v= 331 m/s and f= 256 Hz by rearranging the above expression.

\lambda =\frac{331}{256}

\lambda =1.29 m

Therefore, the wavelength of the wave is 1.29.

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How long does it take a person to skate the width of a hockey rink (85 feet) at a constant speed of 15 feet per second?
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S(travel distance)=85 ft
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-----------------------------------
t (time)=?

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3 years ago
Which of the following units would need to be converted before being used for a calculation
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The units which would need to be converted before being used for a calculation are cm and 0 ks.

<h3>What is Unit?</h3>

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6 0
1 year ago
Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

U=k\frac{q_1q_2}{r_{1,2}}                  (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}           (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J

The electric potential energy between the three charges is 80.91 J

7 0
3 years ago
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