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Fittoniya [83]
2 years ago
7

A light source embedded in a block of glass emits a ray a frequency of 5.7×1014Hz. When the light ray exits the glass into vacuu

m, which of the following statements about what happens will be true? Assume the glass has an index of refraction of 1.5 and is perfectly transparent and does not absorb any light.
A. The speed of the light will increase.
B. The light's amplitude will stay the same.
C. The light's speed will stay the same.
D. The light's amplitude will increase.
E. The light's wavelength will stay the same.
E The light's frequency will decrease.
G. The number of photons will stay the same.
H. The light's frequency will stay the same.
I. The light's wavelength will decrease.
J. The number of photons per second will increase.
K. The light's wavelength will increase.
L. The light's amplitude will decrease.
M. The frequency of the light will increase.
N. The light's speed will decrease.
Physics
1 answer:
Olin [163]2 years ago
7 0

Answer:

Explanation:

Speed of light will increase.

The light's amplitude  will remain constant. provided no energy is absorbed.

The light's wavelength will increase .

The light's frequency will not change.

No of photons per second will remain constant because intensity of light remains constant.

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Positions. Happy to help! Please mark as Brainliest!

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2 years ago
3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
andreev551 [17]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0
2 years ago
An elementary particle of mass m completely absorbs a photon, after which its mass is 1.01m. (a) what was the energy of the inco
sdas [7]
A.) We use the famous equation proposed by Albert Einstein written below:

E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s

Substituting the value:

E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules

b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
8 0
2 years ago
11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0
Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0
2 years ago
What is the depth of the crater if the time is 6.3 s? An astronaut stands by the rim of a crater on the moon, where the accelera
ollegr [7]

Complete Question

An astronaut stands by the rim of a crater on the moon, where the acceleration of gravity is 1.62 m/s2. To determine the depth of the crater, she drops a rock and measures the time it takes for it

to hit the bottom. If the time is 6.3 s, what is the depth of the crater?

Answer:

The depth is 32 m

Explanation:

From the question we are told that

  The  time is  t =  6.3 s

  The acceleration due to gravity is  g =  1.62 \  m/s^2

 

Generally from kinematic equation

    s = ut + \frac{1}{2}  - at^2

Here the u is the initial  velocity and the value is  0 m/s

      s = 0 + \frac{1}{2}  - (1.62) * (6.3)^2

        s = 32 \ m

   

8 0
2 years ago
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