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2 years ago

Which is the BEST example of refraction?

Physics

2 answers:

Mademuasel [1]2 years ago

8 0

Answer:

Option (c) is correct.

Explanation:

When a ray of light passes from one optical medium to another optical medium, the path of ray of light is deviated, this phenomena is called refraction.

It occurs due to the change in speed of light, as it passes from medium to another.

So, when a laser beam passes through a glass block, the refraction takes place.

Aneli [31]2 years ago

6 0

Answer:

c would be your answer just did this on usatest prer

Explanation:

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2 years ago

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2 years ago

Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end

iragen [17]

Answer:

The workdone is $W_d =-4400J$

Explanation:

The free body diagram is shown on the first uploaded image

From the question we are given that

The force is on the force gauge $F = 2750 N$

The distance that Magnus pulled the bus $d = 1.60m$

Generally the workdone by the tension force on Magnus is

$Workdone = Force * displacement \ in \ the \ direction \ of \ force$

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2 years ago

A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fu

Triss [41]

Answer: The answer for A is - v = 786.93 m/s

The answer for B is - v = 122.40 m/s

Explanation:

a) To find the average exhaust speed (v) of the engine we can use the following equation:

F = vΔm

Where:

F: is the thrust by the engine = 5.26 N

Δm: is the mass of the fuel = 12.7 g

Δt: is the time of the burning of fuel = 1.90 s

v = F×ΔT/ΔT

b) To calculate the final velocity of the rocket we need to find the acceleration.

The acceleration (a) can be calculated as follows:

a = F/M

In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:

m = (m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = {2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = 2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g

Now, the acceleration is:

a = 5.26 N/81.65-t 10^³kg} = 64.42 m*s^²

Finally, the final velocity of the rocket can be calculated using the following kinematic equation:

v= v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s

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2 years ago