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2 years ago

What happens to the frequency of a wave if its energy increases?

Physics

2 answers:

aleksandr82 [10.1K]2 years ago

8 0

Answer:

The wavelength decreases as the frequency increases.

Explanation:

The wavelength λ is the distance between successive peaks in a wave.

The frequency f of a wave is the number of complete waves that pass a point in a given time.

Triss [41]2 years ago

7 0

Answer: The frequency of a wave will increase.

Explanation:

The relation between the energy and the frequency is as follows;

$E=h\nu$

Here, h is the planck's constant and $\nu$ is the frequency of the wave.

The energy of the wave is directly proportional to the frequency of the wave. The frequency is inversely proportional to the wavelength. If the energy of the wave increases then the wavelength of the wave decreases.

If the energy of the wave increases then the frequency of the wave will increase.

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Answer:

Car is travelling with 30 m/h.

By unitary method,

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Thus car will travel 60 m in 2 hours.

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2 years ago

standing side by side, you and a friend step off a bridge and fall for 1.6s to the water below. your friend goes first, and you

Andrew [12]

(a) The distance will be more than 2.0 meters.

In fact, you starts your fall after your friend has already fallen 2.0 meters. This means that your friend has already accelerated for a while, therefore his velocity will be greater than yours. But this statement will be actually true for the entire fall, since you has some delay, therefore when your friend will hit the water, the separation between you and him will be greater than the initial separation of 2.0 meters.

b) First of all we need to calculate the height of the bridge with respect to the water. We know that you take 1.6 s to fall down, therefore we can use the following equation:

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(1.6s)^2=12.56 m$

We know that your friend will take 1.6 s to falls down. Instead, you start your jump after he has already fallen 2.0 m, therefore after a time given by the equation:

$S=\frac{1}{2}gt^2$

Using S=2.0 m,

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(2.0 m)}{9.81 m/s^2}}=0.64 s$

So we know that you start your fall 0.64 s after your friend. Therefore, now we can find how much did you fall between the moment you started your fall (0.64 s) and the moment your friend hits the water (1.6 s). Using

$t=1.6 s-0.64 s=0.96 s$

we find

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(0.96 s)^2 =4.52 m$

So, when your friend hits the water, you just covered 4.52 m, while he already covered 12.56 m. Therefore, the separation between you and your friend is more than 2 meters.

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3 years ago

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Georgia [21]

components of the speed of the coin is given as

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$v_x = 6.4 cos60 = 3.2 m/s$

$v_y = vsin60$

$v_y = 6.4 sin60 = 5.54 m/s$

now the time taken by the coin to reach the plate is given by

$t = \frac{\delta x}{v_x}$

$t = \frac{2.1}{3.2}$

$t = 0.656 s$

now in order to find the height

$h = vy * t + \frac{1}{2} at^2$

$h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2$

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