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Lelu [443]
3 years ago
13

What happens to light waves at the interface between different media?

Physics
1 answer:
nevsk [136]3 years ago
6 0

Answer:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Explanation:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Hope this helps.

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Nicolaus Copernicus was considered the <span>"Father of Astronomy" because he believed the earth was the center of the solar system. The rest of the choices do not answer the question above.</span>
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4 years ago
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The length of second hand of clock is 14cm, an ant sits on the top of second hand. find the following
Anastaziya [24]

Answer:

The answer is below

Explanation:

i) Since the length of the second clock (radius) is 14 cm = 0.14 m, the distance covered by the second hand in one revelution is:

Distance covered = 2πr = 2π(0.14) = 0.88 m

The time taking to complete one revolution = 60 seconds, hence;

Speed = distance covered in one revolution / time take o complete a revolution

Speed = 0.88 m / 60 s = 0.0147 m/s

ii) Distance covered in 150 s = speed * 150 s = 0.0147 * 150 = 2.2 m

iii) Displacement in 150 seconds = distance from initial position to final position

At 150 s, the hand has covered 2 revolutions and moved 30 s. Hence:

Displacement in 150 seconds = speed * 30 s = 0.0147 * 30 = 0.44 m

4 0
3 years ago
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib
Natasha_Volkova [10]

Answer:

F = 1958.4 N

Explanation:

By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side

so we have

L_1A_1 = L_2A_2

1.20(\pi 18^2) = L_2(\pi 5^2)

L_2 = 15.55 m

so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m

So here the net pressure on the smaller area is given as

P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)

excess pressure exerted on the smaller area is given as

P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)

P_{ex} = 2.49\times 10^5 Pascal

now the force required on the other side is given as

F = P_ex (area)

F = (2.49 \times 10^5)(\pi (0.05)^2)

F = 1958.4 N

3 0
4 years ago
A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and
iogann1982 [59]

Answer:

p_2 = 1.76 atm

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}

Putting the values

\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}

9.733*10^{-3} = \frac{p_2 *1.68}{304}

9.733*10^{-3}*304 = p_2*1.68

\frac{9.733*10^{-3}*304}{1.68} =p_2

p_2= 1.76 atm

7 0
4 years ago
A 47.5 g block of an unknown metal is heated in a hot water bath to 100.0
lozanna [386]
The answer might be B i hope is helps
3 0
4 years ago
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