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1 year ago

How does light behaves when light passes through water?

Physics

1 answer:

MrMuchimi1 year ago

7 0

light bends towards the normal when passing from air to water.

Explanation:

when Light passes from air to water It gets refracted and bend towards normal and if it passes from water to air it bends away from normal. This is because of the difference in density of the medium. Light when passing through higher density to lower bends away from normal and on passing through lower density to higher bends towards normal.

in general, some of the rays of light will reflect from the surface of water but other will pass through the water. The normal is the perpendicular line drawn on the surface at the exact point of incident.

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Dmitry [639]

Answer:

The arrow rise 44.64 m high assuming air friction is negligible

Explanation:

According to the law of conservation of energy

Kinetic energy = Potential energy at highest

$\frac{1}{2}kx^2=mgh$ ----1

We know that F=kx

So, $k = \frac{F}{x}\\k=\frac{175}{0.75}$

Substitute the value in 1

$\frac{1}{2}(\frac{175}{0.75})(0.75)^2=0.15 \times 9.8 \times h\\\frac{\frac{1}{2}(\frac{175}{0.75})(0.75)^2}{0.15 \times 9.8}=h\\44.64 = h$

Hence the arrow rise 44.64 m high assuming air friction is negligible

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3 years ago

A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met

Tanya [424]

option E

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2 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra

Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

$m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}$

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

$v=\sqrt{\frac{GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

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$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula,

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B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s$

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Explanation:

<u> Please refer to the attachment for answers and explanation</u>

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