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goldenfox [79]
2 years ago
6

A beam of monochromatic light with a wavelength of 400 nm in air travels into water. what is the wavelength of the light in wate

r? give your answer in nanometers.
Physics
1 answer:
slava [35]2 years ago
5 0
The refractive index of water is n=1.33. This means that the speed of the light in the water is:
v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.33 }=2.26 \cdot 10^8 m/s

The relationship between frequency f and wavelength \lambda of a wave is given by:
\lambda= \frac{v}{f}
where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water \lambda_w to that in air \lambda as
\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}
where v is the speed of light in water and c is the speed of light in air. Re-arranging this formula and by using \lambda=400 nm, we find
\lambda_w = \lambda \frac{v}{c}=(400 nm) \frac{2.26 \cdot 10^8 m/s}{3 \cdot 10^8 m/s}=301 nm
which is the wavelength of light in water.
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What if is a forecasting game, below are actions on your document and all you have to do is predict what will happen after the a
Talja [164]

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If you press one of the arrow keys with the Shift key pressed, you can select a section of text.

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7 0
2 years ago
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When popcorn is heated over a flame, the kernel burst open. why does this occur?
Bezzdna [24]
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4 0
3 years ago
A 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal exp
Gnom [1K]

Answer:

a.) 1567.2 m/s

b.) 149.4 m/s

Explanation:

Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.

The x-component of the third part can be calculated by assuming that it moves in a positive x axis.

The third mass = 26 - ( 7.8 + 8.8)

The third mass = 26 - 16.6

The third mass = 9.4kg

since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion

26 x 350 = -8.8 x 640 + 9.4V

9100 = -5632 + 9.4V

9.4V = 9100 + 5632

9.4V = 14732

V = 14732/9.4

V = 1567.2 m/s

(b) y-component of the velocity of the third part will be

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V = 1404/9.4

V = 149.4 m/s

7 0
3 years ago
gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure
Mamont248 [21]

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be '\text{P}_2'.

As we know that Boyle's law formula states that;

                    P_1 \times V_1 = P_2 \times V_2

where, P_1 = original pressure of gas in the container = 310 mm hg

           P_2 = required new pressure

            V_1 = volume of gas in the container = 185 ml

            V_2 = desired new volume of the gas = 74 ml

So,  P_2 = \frac{P_1 \times V_1}{V_2}  

       P_2 = \frac{310 \times 185}{74}

            =  775 mm hg

Hence, the required new pressure is 775 mm hg.

7 0
3 years ago
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