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Ainat [17]
3 years ago
7

A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surfac

e would the letters appear to be when viewed from directly above?
Physics
1 answer:
7nadin3 [17]3 years ago
8 0

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

i.e.

Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

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A 7-kg bowling ball moving at 4 m/s strikes a 1 kg bowling pin. If the ball slows to 2 m/s in 0.05 s, how much force does it exe
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You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, w
solniwko [45]

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

                                   = 0.025×< -2, 0, -7 >

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Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

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                                    = < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact  =  < -0.05,  0,  0.175 > + < 0.1, 0.1, -0.075>

                                                      = < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together  after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity =  v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

          = < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

7 0
3 years ago
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