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dem82 [27]
3 years ago
8

A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?

Physics
1 answer:
posledela3 years ago
6 0

Answer:

118\; \rm Hz.

Explanation:

The frequency f of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of v= 295\; \rm m\cdot s^{-1}. In other words, the wave would have traveled 295\; \rm m in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that 295\; \rm m? The wavelength of this wave\lambda = 2.50\; \rm m gives the length of one wave cycle. Therefore:

\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118.

That is: there are 118 wave cycles in 295\; \rm m of this wave.

On the other hand, Because that 295\; \rm m of this wave goes through that point in each second, that 118 wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

f = 118\; \rm s^{-1} = 118\; \rm Hz.

The calculations above can be expressed with the formula:

\displaystyle f = \frac{v}{\lambda},

where

  • v represents the speed of this wave, and
  • \lambda represents the wavelength of this wave.

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timofeeve [1]

The solution would be like this for this specific problem:

 

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0
3 years ago
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Nataly_w [17]

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Explanation:

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The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by t
stich3 [128]

Answer:

\dfrac{dV}{dt} = -0.466 m^3/s

Explanation:

given,

P (in kilo pascals),         volume V (in liters),        temperature T (in kelvins)

P V = 8.31 T

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Rate at which volume is changing = ?

V = 8.31 \dfrac{T}{P}

\dfrac{dV}{dt} = 8.31 \dfrac{P\dfrac{dT}{dt}-T\dfrac{dP}{dt}}{P^2}

\dfrac{dV}{dt} = 8.31 \dfrac{18 \times 0.1-285\times 0.07}{18^2}

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nasty-shy [4]
F=g(m1*m2)/r^2
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7 0
3 years ago
Read 2 more answers
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