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svlad2 [7]
4 years ago
7

As the wavelength of an electromagnetic wave ___ the frequency of the wave ________.

Physics
1 answer:
adoni [48]4 years ago
8 0

Answer:

As the wavelength of an electromagnetic wave _decrease__ the frequency of the wave _increase_______.

Explanation:

What is the relationship between frequency and wavelength?

Wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.

That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. The equation that relates wavelength and frequency is:

V= fλ

where v= velocity

           f= frequency

            λ = wavelength

⇒ f = v/λ  

also f ∝ 1/λ

For electromagnetic radiation, the speed is equal to the speed of light, c, and the equation becomes:

C= fλ

where c= Speed of light

           f= frequency

            λ = wavelength

⇒ f = v/λ  

also f ∝ 1/λ

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Explanation:

3 0
3 years ago
A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while
rjkz [21]

To solve this problem it is necessary to apply Boyle's law in which it is specified that

P_1V_1 =P_2 V_2

Where,

P_1 and V_1 are the initial pressure and volume values

P_2 and V_2 are the final pressure volume values

The final pressure here is the atmosphere, then

P_2 = 101325 \approx 1*10^5Pa

h = 10m

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V_1 = 3.0L

Pressure at the water is given by,

P_1 = P_2 -\rho gh

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Using Boyle equation we have,

V_2 = \frac{P_1V_1}{P_2}

V_2 = \frac{198000*3*10^5}{10^5}

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4 0
3 years ago
Jim and Sally both do identical jobs. Jim works quickly while Sally works slowly. Which of the following is true?
Lynna [10]
The answer should be C
5 0
3 years ago
The function of the __________is to store and supply fuel to the cylinder chamber where it can be mixed with air, vaporized, and
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hope it helps (^^)

# Cary on learning

3 0
2 years ago
1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is
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Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N

4 0
3 years ago
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