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Naddika [18.5K]
3 years ago
14

"when light travels from air into a piece of glass, a change in direction may occur. what is the name of this phenomenon

Physics
2 answers:
Kaylis [27]3 years ago
8 0
As light travels through a medium, it moves in a straight line path,but when light passes from one medium into another medium, the light path bends,Refraction takes place. 
jek_recluse [69]3 years ago
5 0
Glass is a denser medium(medium in which particles are close to each other) whereas air is rarer medium(medium in which particles are far away from each other). Whenever light rays travel from rarer to denser medium, speed of light decreases. This is because, in denser medium, light rays interact with more particles and due to this there is a time delay. Geometrically,When light rays enter glass or any other denser medium, it bends towards the normal of denser medium because refractive index of glass is more than that of air. Finally, when light ray travels from glass to air, the emergent ray will be parallel(it is not exactly parallel and this is due to curvature of space-time. But the deviation is negligible in all the geometrical optiks experiment) to the incident ray because light ray travels in the same medium i.e air.

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Cryosphere is the sphere that contains all of the frozen water on Earth. The correct option among all the options that are given in the question is the fourth option or the last option. Cryosphere is actually those parts of the surface of the earth where the water is in a frozen state, like in seas and oceans.
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3 years ago
You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi
Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0
3 years ago
I need help can someone help me pls
pychu [463]

Answer:

copper

Explanation:

4 0
3 years ago
Read 2 more answers
Imagine you have just witnessed a small avalanche on a mountain while skiing, and two slushy snowballs just crashed together in
inn [45]
We have that the momentum p is given by the formula p=mv where m is the mass and v is the velocity. Since for A p=-14kgm/s and m=7, we have that the velocity is -14/7=-2m/s. Hence its speed is 2 m/s.
For b we have that p=15kgm/s and v=3m/s. Because m=p/v, we have m=3kg.
We also have that the momentum is conserved in this system. Hence, the net sum of the momentum of the 2 snowballs equals the momentum of the single giant ball. Hence, p(total)=p(combined)=-14+15=1kgm/s (momentum is a vector; the positive sign means that it tends to the positive direction).

6 0
3 years ago
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Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro
blagie [28]

Answer:

21.2\times 10^{-6} T

Explanation:

i  = magnitude of current in each wire = 2.0 A

a  = length of the side of the square = 4 cm = 0.04 m

r  = length of the diagonal of the square = \sqrt{2} a = \sqrt{2} (0.04) = 0.057 m

B = magnitude of magnetic field by wires at A and C

B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )

B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )

B = 10\times 10^{-6} T

B' = magnitude of magnetic field by wire at B

B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )

B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )

B' = 7.02\times 10^{-6} T

Net magnitude of the magnetic field at D is given as

B_{net} = \sqrt{B^{2}+B^{2}} + B'

B_{net} = \sqrt{2} B + B'

B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})

B_{net} = 21.2\times 10^{-6} T

8 0
3 years ago
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