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Yuri [45]
4 years ago
13

A point source of light is submerged 3.3 m below the surface of a lake and emits rays in all directions. On the surface of the l

ake, directly above the source, the area illuminated is a circle. What is the maximum radius that this circle could have? Take the refraction index of water to be 1.333.
Physics
1 answer:
krok68 [10]4 years ago
8 0

Answer:

Maximum Radius = 2.89m

Explanation:

The maximum radius will be determined by the angle of incidence which is equal to the critical angle. Now, any angle larger than that will make the light to be totally internally reflected. Hence, we can figure out that angle from Snell’s law where the refracted angle is 90°, and then use the tangent function.

From Snell's law;

n_air*sin90° = n_water*sin(θ _c)

Where;

θ_c is the critical angle

Refractive index of water; n_water = 1.333

Refractive index of air;n_air = 1

Thus;

1*1 = 1.33sinθ_c

sinθ_c = 1/1.33

θ_c = sin^(-1)0.7519

θ_c = 48.76°

Like I said earlier, we'll use tangent to find the radius.

Thus;

tanθ_c = d/R

From the question, d = 3.3m

Thus;

3.3/tan48.76 = R

So, R = 2.89m

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KengaRu [80]

Answer:

V = -1.33 m/s

Explanation:

Given,

The mass of the camper, m = 100 kg

The velocity of the camper, v = 3.0 m/s

The combined mass of canoe and another camper, M = 225 kg

The velocity of the combined canoe and another camper, V = ?

According to the law of conservation of momentum,

                          MV + mv = 0  (Since the initial momentum of the dock is 0)

                             V = -mv/M

                                =  -100 x 3/225

                                = -1.33 m/s

The negative sign indicates that the combined objects moves opposite to that of the camper.

Hence, the velocity of the combined canoe and camper is, V = -1.33 m/s

6 0
4 years ago
Show that the relationship between impulse and the change in momentum is another way of stating Newton's second law of motion.
user100 [1]
My answer to this question.

6 0
3 years ago
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VikaD [51]

Answer:

Circular wave

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A wave that propagates in circular form on the surface of water falls into this category.

6 0
3 years ago
Which factor affects elastic potential energy but not gravitational potential energy?
julsineya [31]

Explanation:

The energy stored in an elastic objects as a result of deformation is called elastic potential energy. The energy stored in a spring is given by :

E=\dfrac{1}{2}kx^2

Where

k = spring constant

x = compression or stretching in an spring

While gravitational potential energy is given by :

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