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vekshin1
3 years ago
13

What is meant by the term 'total internal reflection'? (GCSE Level)

Physics
1 answer:
scZoUnD [109]3 years ago
6 0

Explanation:

Total Internal Reflection (TIR) is a phenomenon in optics, by which light experiences complete reflection at an interface between two media.

In the figure i > © experience total internal reflection

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What is the force on an object that goes from 35 m/s to 85 m/s in 20 seconds and has a mass of 148 kg
Sever21 [200]
F=ma
F = 148×(85-35)÷20
F = 148×(50÷20)
F = 148×2.5
F = 370N
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Reading glasses use what property of light waves
tresset_1 [31]
Visible range of electromagnetic radiation
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For all turning vehicles, the rear wheels follow a ______ than the front wheels.
aliya0001 [1]

Answer:

Shorter path

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For all turning vehicles, the rear wheels follow <u>Shorter path</u> than the front wheels.

Any turning vehicle, the rear(the back part of something, especially a vehicle.) wheels follow a shorter path than the front wheels. The longer the vehicle is, the greater the difference will be in path. Trucks initially swing out before making a turn

3 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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