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Scilla [17]
3 years ago
9

A Ray of light in air makes an angle of 45 degree at the surface of a sheet of ice . the ray is refracted within the Ice at an a

ngle of 30 degree . a) what is the critical angle for the Ice? b) A speck of dirt is embedded 2CM below the surface of the Ice . what is the Apparent depth when viewed at normal incidence​
Physics
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

Explanation:

a )

Refractive index of ice = sini / sin r where i is angle of incidence and r is angle of refraction

angle of incidence = 90 - 45 = 45

angle of refraction = 30

refractive index of ice   μ  = sin 45 / sin 30

= 2/ √2

μ  = √2 = 1.414

μ  = 1 / sin C where C is critical angle for ice - air interface

sinC = 1 / μ

= 1 / 1.414

= .70

C = 44 .42° .

b )

real depth / apparent depth = refractive index

2 / apparent depth = 1.414

apparent depth = 2 / 1.414

= 1.414 cm .

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A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
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Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

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The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

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N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0
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