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Sunny_sXe [5.5K]
3 years ago
14

A wave is traveling at a speed of 15 m/s and it's wavelength is 5 m. Calculate the waves frequency

Physics
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer:

The frequency of this wave is 3\; \rm Hz.

Explanation:

The frequency f of a wave is the number of wavelengths that this wave covers in unit time (typically a second.)

The wave in this question travels at v = 15\; \rm m \cdot s^{-1}. In other words, this wave covers 15\; \rm m in unit time (a second.) How many wavelengths \lambda would that 15\; \rm m\; correspond to?

The question states that the wavelength of this wave is \lambda = 5\; \rm m. Therefore, there would be 15 / 5 = 3 wavelengths in the 15\; \rm m span that this wave covered in the unit time of one second (1\; \rm s.) Hence, the frequency of this wave would be 3\; \rm s^{-1} (three per second,) which is equivalent to 3\; \rm Hz (three Hertzs.)

In general, the frequency f of a wave with speed v and wavelength \lambda would be:

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

For the wave in this question:

\begin{aligned}f &= \frac{v}{\lambda} \\ &= \frac{15\; \rm m \cdot s^{-1}}{3\; \rm s} = 3\; \rm s^{-1} = 3\; \rm Hz\end{aligned}.

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The light can definitely change the mystery material. This can occur through a change in temperature or color (option C).

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Moreover, this phenomenon can lead to two main changes:

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Note: This question is incomplete; here is the missing part:

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2 years ago
Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins
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Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

X(t) = 4t^2 - 3t + 40 ...... (1)

For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

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60t - 15 = 0

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Total displacement will be

X(t) = 4(0.25)^2 - 3(0.25) + 40

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Total displacement = 39.5 m

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V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

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