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Anna71 [15]
3 years ago
5

How can you double the frequency of a wave if you have control over both the wavelength and the wave velocity?

Physics
2 answers:
iren2701 [21]3 years ago
6 0

Answer:

Either double the speed or reduce the wavelength to half

Explanation:

We know that velocity of the wave is given by velocity=wavelength ×frequency

v=\lambda f

So f=\frac{velocity}{wavelength}=\frac{v}{\lambda }

As in the question it is given that we have control on both velocity and wavelength

So we have to double the the frequency then we can double the speed or reduce the wavelength to half

sammy [17]3 years ago
5 0

Answer:

Change the wavelength to half while keeping velocity constant or change the wave velocity to 2 times  while keeping wavelength constant.

Explanation:

The frequency of a wave can be defined as the rate of change of wave speed with respect to the wavelength of the wave.

Mathematically,

f=\frac{v}{\lambda}

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

If the observer want to double the frequency of the wave, then he should increase the wave velocity two times and take wavelength constant for this case or either he should half the wavelength of the wave to take the velocity of the wave constant.

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C is the correct answer.

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ikadub [295]

Answer:

a) a_{1}=3.7 m/s^{2}

b) a_{2}=3.68 m/s^{2}

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

v_{f}^{2}=v_{i}^{2}+2a\Delta x

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positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}

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a_{2}=3.68 m/s^{2}

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I hope it helps you!

8 0
4 years ago
A supertanker filled with oil has a total mass of 6.1 x108 kg. If the dimensions of the ship are those of a rectangular box 300
IrinaVladis [17]

Answer:

The bottom of the sea is 25 m below sea level.

Explanation:

Given data

Mass = 6.1 × 10^{8} \ kg

\rho_{sea} = 1020\  \frac{kg}{m^{3} }

We know that Buoyant force on the tank is equal to gravity force of the tank.

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(\rho_{Fluid})  (V_{disp}) = m

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V_{disp} = 598039.21 m^{3}

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Therefore the bottom of the sea is 25 m below sea level.

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