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3 years ago

What are two uses of total internal reflection? (GCSE Physics)

Physics

1 answer:

jeka943 years ago

3 0

Answer:

1) Used in optical instruments such as telescopes.

2) Used to form mirages

Explanation:

1) Perhaps the simplest example of this is the astronomical refractor telescope with a right-angle eyepiece holder. Astro scopes mostly point at things high in the sky, but it is inconvenient to place your eye low to look up through the scope, so the light path is bent 90 degrees just before the eyepiece. This can be done with a mirror, but using a simple 45 degree prism (internal angles 45, 45, and 90 degrees) will do the 90 degree bend more efficiently.

2) So when a light pass from cold air to hot air light tends to bend from its path which is known as refraction. As the light get refracted it reaches to a point where the light tends to form 90 degree angle.

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Which of the following is a process that occurs at divergent boundaries?

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3 years ago

In what region of the electromagnetic spectrum is a photon found that possesses twice as much energy as one in the blue region (

iren2701 [21]

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3 years ago

What happens to a day of light that slows down when it hits a new medium at an angle

Anika [276]

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3 years ago

Two small nonconducting spheres have a total charge of 93.0 μC . Part A

Travka [436]

Answer:

charge on each

Q1 = 2.06 × $10^{-5}$ C

Q2 = 7.23 × $10^{-5}$ C

when force were attractive

Q1 = 1.07 × $10^{-4}$ C

Q2 = -1.39 × $10^{-5}$ C

Explanation:

given data

total charge = 93.0 μC

apart distance r = 1.14 m

force exerted F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC = 93 × $10^{-6}$ C

and

According to Coulomb's law force between two sphere is

Force F = $\frac{K Q1 Q2 }{r^2}$ .........1

Q1Q2 = $\frac{F*r^2}{k}$

here F is force and r is apart distance and k is 9 × $10^{9}$ N-m²/C² put all value we get

Q1Q2 = $\frac{ 10.3*1.14^2}{9*10^9}$

Q1Q2 = 1.49 × $10^{-9}$ C²

and

we have Q2 = 93 × $10^{-6}$ C - Q1

put here value

Q1² - 93 × $10^{-6}$ Q1 + 1.49 × $10^{-9}$ = 0

solve we get

Q1 = 2.06 × $10^{-5}$ C

and

Q1Q2 = 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = 1.49 × $10^{-9}$

Q2 = 7.23 × $10^{-5}$ C

and

if force is attractive we get here

Q1Q2 = - 1.49 × $10^{-9}$ C²

then

Q1² - 93 × $10^{-6}$ Q1 - 1.49 × $10^{-9}$ = 0

we get here

Q1 = 1.07 × $10^{-4}$ C

and

Q1Q2 = - 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = - 1.49 × $10^{-9}$

Q2 = -1.39 × $10^{-5}$ C

5 0

3 years ago

Help!! ASAP plz

VashaNatasha [74]

Can you include an image of the object and it’s dimensions?

4 0

3 years ago