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JulijaS [17]
3 years ago
12

When light moves from air to glass, part of the light is reflected and part of it is refracted. In the image, which ray shows re

fraction?

Physics
2 answers:
marissa [1.9K]3 years ago
6 0
You have not provided the diagram, therefore, I cannot provide an exact answer.
However, I will try to help by explaining how to solve this problem.

When light moves from air to glass:
1- part of the light is reflected back into the air where the angle of incidence is equal to the angle of reflection
2- part of the light enters the water and refracts. The angle of refraction can be calculated using Snell's law.

In a diagram, the reflected ray would be the one getting back into air while the refracted ray would be the one entering the water.

You can check the attached diagram for further illustrations.

Hope this helps :)

lesya [120]3 years ago
4 0

the 2nd one

(when the light moves from air to glass)

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75

Explanation:

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A modified Atwood machine is at rest. The hanging block has a mass of 3kg. The black on wheels has unknown mass M1. When release
fgiga [73]

Answer:

From the figure,

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7 0
3 years ago
A 8.10×10^3 ‑kg car is travelling at 25.8 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 3.9
KengaRu [80]

Answer:

The magnitude of the average drag force is 2412.34 N.

Explanation:

Given that,

Mass of car m=8.10\times10^{-3}\ kg

Velocity v = 25.8 m/s

Distance d= 3.90\times10^{2}

Speed of car = 13.1 m/s

Height = 12.5 m

We need to calculate the magnitude of the average drag force

Using equation kinetic energy

K.E_{i}=K.E_{f}+P.E+F_{d}

\dfrac{1}{2}mv_{i}^2=\dfrac{}{}mv_{f}^2+mgh+F\times d

Where, v_{i} = initial velocity

v_{f} = final velocity

h = height

g = acceleration due to gravity

F_{d}=drag force

m = mass of the car

d = distance

Put the value into the formula

\dfrac{1}{2}\times8.10\times10^{3}\times25.8=\dfrac{1}{2}\times8.10\times10^{3}\times13.1+8.10\times10^{3}\times9.8\times12.5+F\times3.90\times10^{2}

F=\dfrac{\dfrac{1}{2}\times8.10\times10^{3}\times25.8-\dfrac{1}{2}\times8.10\times10^{3}\times13.1-8.10\times10^{3}\times9.8\times12.5}{3.90\times10^{2}}

F=-2412.34\ N

|F|=2412.34\ N

Hence, The magnitude of the average drag force is 2412.34 N.

4 0
4 years ago
A vehicle approaches a set of traffic lights with a speed of 15m/s. When 50m from the lights the driver brakes and comes to a re
sattari [20]

Answer:

–2.25 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 15 m/s

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Final velocity (v) = 0 m/s

Deceleration (a) =?

v² = u² + 2as

0² = 15² + (2 × a × 50)

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Collect like terms

0 – 225 = 100a

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Divide both side by 100

a = –225/100

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Thus, the deceleration of the vehicle is –2.25 m/s²

3 0
3 years ago
A lens is used to make an image of magnification -1.50. The image is 30.0 cm from the lens. What is the object’s distance (unit=
photoshop1234 [79]

Answer:

The object distance is 20cm

Explanation:

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Required

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Multiply both sides by -1

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Divide through by 1.50

1.50U/1.50 = 30/1.50

U = 30/1.50

U = 20cm

Recall that U represented the object distance.

Hence, the object distance is 20cm

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