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Genrish500 [490]

3 years ago

10

Light travels through water at 2.2506 x 108 m/s. A person shines a light at a friend that is underwater. If the ray in the water

makes an angle of 38° with the normal, what is the angle of incidence?

2 answers:

Andrew [12]3 years ago

8 0

please find your answer in the attached image, than you for 48 points.

vagabundo [1.1K]3 years ago

5 0

Answer:

I WILL TRY AND IF I SOLVE IT I WILL SEND YOU THE IMAGE..

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The atomic number of an atom is the number of ____.

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Suppose you are in total equilibrium in water (in other words you are floating and not moving in any direction)What could you do

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Explanation:

According to newtons first law of motion:

'' a body will continue in its state of rest or uniform motion along a path unless it is acted upon by an external force".

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3 years ago

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, $d=2.5\times 10^{-15}\ m$

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Let t is the time interval required for the strong interaction to occur. The speed is given by :

$c=\dfrac{d}{t}$

$t=\dfrac{d}{c}$

$t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}$

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3 years ago

Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h

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Disagree.
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4 0

2 years ago

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A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

vladimir2022 [97]

Answer:

a)KE=878.8 J

b)W=2636.4 J

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

a)

We know that kinetic energy KE is given as follows

$KE=\dfrac{1}{2}mu^2$

m=mass

u=velocity

Now by putting the values in the above equation we get

$KE=\dfrac{1}{2}\times 65\times 5.2^2\ J$

KE=878.8 J

b)

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Now by putting the values in the above equation we get

$W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J$

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

8 0

3 years ago