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3 years ago

((PLEASE HELP))

Physics

2 answers:

Degger [83]3 years ago

5 0

Answer:

Angle of refraction

Explanation:

The incident ray is the ray before it reaches the surface.

The refracted ray is the ray after it reaches the surface.

n₁ is called the index of incidence.

n₂ is called the index of refraction.

θ₁ is called the angle of incidence.

θ₂ is called the angle of refraction.

They are related by Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

il63 [147K]3 years ago

5 0

Answer:

Angle of refraction

Explanation:

The angle θ2 represents the angle of refraction.

Refraction is the change in direction of light rays as it passes from one medium to another. As light rays passes from one medium to another, it changes direction.

If the incident light is passing from the less dense medium to denser medium, the refracted ray will bend towards the normal. In this case the angle of refraction 'r' is always less than the angle if incidence. (According to the diagram)

Similarly, If the incident light is passing from the denser medium to less dense medium, the refracted ray will bend away from the normal. In this case the angle of refraction 'r' is always greater than the angle of incidence.

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Which of the following is proportional to the net external force acting on a body?

swat32

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You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is

madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book $(m_0)=1.7 kg$

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time taken=0.64 s

$h=ut+frac{at^2}{2}$

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$\alpha =93.84 rad/s^2$

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$T=m(g-a)$

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T=8.364 N

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3 years ago

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3 years ago

A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is

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Hello!

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$P = i^2R$

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2 years ago

A 66-kg diver jumps off a 9.7-m tower. (a) Find the diver's velocity when he hits the water. (b) The diver comes to a stop 2.0 m

Mariana [72]

Answer:

(a) 13.795 m/s.

(b) -3140.28 N.

Explanation:

(a) Using newton's equation of motion,

v² = u² + 2gs.......................... Equation 1

Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.

Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².

Substitute into equation 1

v² = 0² + 2×9.81×9.7

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v = √(190.314)

v = 13.795 m/s.

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(b)

F = ma.................... Equation 2

Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.

Also using

v² = u² + 2as ............ Equation 3

Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s

Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s

Substitute into equation 3

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a = 190.30203/-4

a = -47.58 m/s²

Also given: m = 66 kg,

Substitute into equation 3

F = (-47.58)(66)

F = -3140.28

Note: The Force is negative because it act against the motion of the diver.

Hence the net force exerted on the diver while in the water = -3140.28 N.

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