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2 years ago

What is the frequency of an electromagnetic wave that has a wavelength of 300,000 km? (the speed of light is 300,000 km/s.)?

1 answer:

6 0

Frequency represents the number of complete oscillations in one second. it is measured in Hertz (Hz). Electromagnetic waves are waves which do not require a material media for transmission. They travel with a speed of light.
The speed (m/s) of a wave is given by frequency (Hz) × Wavelength (m)
Speed is 300,000 km/sec or 300,000,000 m/s and the wavelength is 300,000 km or 300,000,000 m.
Frequency = speed÷ wavelength
= 300000000 ÷ 300000000 = 1
Therefore, the frequency of the wave is 1Hz

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3 years ago

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2 years ago

Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, $\overrightarrow{v_1}=-120\ \vec{i}$

Velocity in north direction is, $\overrightarrow{v_2}=120\ \vec{j}$

Now, since $v_1\ and\ v_2$ are perpendicular to each other, their resultant magnitude is given as:

$|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}$

Plug in the given values and solve for the magnitude of the resultant.This gives,

$|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h$

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

$x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)$

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

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2 years ago

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3 years ago

A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the

ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

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When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

$\frac {Vf^{2}-Vo^2}{2.a} =X$