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stepladder [879]
2 years ago
6

Explain the process of refraction of light​

Physics
1 answer:
brilliants [131]2 years ago
4 0

The process of refraction of light​ occurs when light rays bends when travelling between media of different densities.

What is refraction of light?

Refraction of light is the bending of light rays or the change in the direction of light rays as it travels between media of different densities.

Light waves travel faster in media of less density than media of more density.

The change in density of the media makes light waves to be bend towards or away from the normal.

For example, when light travels from less dense air to more dense water, the light rays are bent towards the normal. However, when light rays travel from water to air, the light rays are refracted away from the normal.

In conclusion, refraction of light waves occur when light crosses the boundary of media of different densities.

Learn more about refraction of light at: brainly.com/question/27932095

#SPJ1

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If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?
aivan3 [116]

Initial velocity = \(v_0\)

acceleration in the downward direction = -9.8 \(\frac {m}{s^2}\)

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

\(v^2 = {v_0}^{2} + 2aH

\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410

\({v_0}^{2} = 2 \times 9.8 \times 0.410\)

\(v_0 = 2.834 \frac {m}{s}\)

Speed with which the flea jumps = \(2.834 \frac {m}{s}\)

4 0
4 years ago
Think of one example where Bob would need to calculate the net force on a person at the water park.
gregori [183]
Well Bob would need to calculate to net force of someone going down a water slide. Since the person is going down the slide, the person will go faster, depending on their mass/weight and the gravitational pull. As phrased in Newton’s Second Law.


Newton’s Second Law:
Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
7 0
3 years ago
A football is thrown at an angle of 30.° above the horizontal. The magnitude of the horizontal
VladimirAG [237]

Answer:

1.53 s

Explanation:

Initially vertical component of velocity of the ball, uy = 7.5 m/s

Net displacement is vertical direction is zero, Δy =0

Use second equation of motion:

Δy = uy t + 0.5 a t²

Here, acceleration a = -g                           (g =9.8 m/s²)

Substitute all the values and solve for g

0 = 7.5 t -0.5 (9.8)t²

7.5 t = 4.9 t²

t = 1.53 s

8 0
3 years ago
Read 2 more answers
A train travels 400km in 2 hours. what is its velocity in km/hr?
Alex_Xolod [135]
Velocity = displacement / time
v = 400 / 2
v = 200 km/h

In short, Your Answer would be 200 km/h

Hope this helps!
7 0
4 years ago
Read 2 more answers
You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 mm/???????? passes you. When the rear of the bu
neonofarm [45]

Answer:

You run 74.1409 mm and you are running at 11.9311 mm/s

Explanation:

If the bus is moving at a constant speed of 5.00mm/s and you start to run when the bus pass you by 12 mm, the equation that describe the position of the bus is:

Xb = 12.0 mm + (5.00 mm/s)*t

Where t is the time in seconds.

If you start to run toward it with a constant acceleration of 0.960 mm/s2, the equation that describe your position is:

X_y=\frac{1}{2} (0.960\frac{mm}{s^{2}})*t^{2}

So, the time t when you catch up the rear of the bus is the time when Xb is equal to Xy. This is:

X_b=X_y\\12+5t=\frac{1}{2} 0.960t^{2} \\0.48t^{2}-5t-12=0

Then, solving the quadratic equation, we obtain that t is equal to 12.4282 s

So, if we replace this value of t in the equation of Xy, we obtain how far you have run before you catch up with the rear of the bus. This is:

X_y=\frac{1}{2} (0.960\frac{mm}{s^{2}})*12.4282^{2}

Xy = 74.1409 mm

Then, the equation of your velocity a time t can be write as:

Vy=0.960\frac{mm}{s^{2} }*t

So, the velocity when you catch up the rear of the bus is:

Vy=0.960\frac{mm}{s^{2} }*12.4282s

Vy = 11.9311 mm/s

3 0
3 years ago
Read 2 more answers
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