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3 years ago

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The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave?

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?

<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>

<h3>Given:-</h3>

Velocity (V) = 40 m/s

Wavelength $(\lambda)$ = 80 cm = 0.8 m

<h3>To Find:-</h3>

The frequency (F) of the wave.

<h2>Solution:-</h2>

We know,

$\bf V \: = \: F \: × \: \lambda$

40 = F × 0.8

F = $\frac{40}{0.8}$

F = 50

<h3>The frequency of the wave is <u>5</u><u>0</u><u> </u><u>H</u><u>z</u>. [Answer]</h3>

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Any object that is not in the floor or resting in some site is being affected by the gravitational force (remember that we are ignoring air resistance)

Then the acceleration of this object will be equal to the gravitational acceleration:

a = -9.8m/s^2

Where the minus sign is because this acceleration goes down.

To get the velocity equation we need to integrate over time, we will get:

v(t) = ( -9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity.

To get the position equation we need to integrate over time again, we will get:

p(t) = (1/2)*( -9.8m/s^2)*t^2 + v0*t + H

Where H is the initial height.

p(t) = (-4.9 m/s^2)*t^2 + v0*t + H

The object will hit the ground when p(t) = 0

Then we need to solve for t the next equation:

(-4.9 m/s^2)*t^2 + v0*t + H = 0

Notice that the only things we need to know are:

H = initial height (we know that is the same for both bullets)

v0 = initial vertical velocity (also is the same for both bullets)

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