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SVETLANKA909090 [29]
3 years ago
7

A person uses eyeglasses with a power of -5.5 D. The lenses are 1.8 cm from the person's eye. What is the furthest this person c

an see clearly to when not wearing their eyeglasses?
A. 2.7 cm
B. 20 cm
C. 28 cm
D. 35 cm
E. 52 cm
Physics
1 answer:
bija089 [108]3 years ago
6 0

Answer:

The correct answer is option B.

Explanation:

power(P)=\frac{1}{f}

f = focal length

P = -5.5 D (negative power means that lens is concave)

f=\frac{1}{P}=\frac{1}{-5.5 D}= -0.1818 m

1 m = 100 cm

f = -0.1818 m = -0.1818 × 100 cm= -18.18 cm

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

u=\infty

\frac{1}{-18.18 cm}=\frac{1}{\infty }+\frac{1}{v}

v = -18.18 cm

v = 1.8 cm - far point

far point = 1.8 cm - v =1.8 cm - (-18.18 cm) =  19.98 cm

But closest answer from the given options is option B.Hence, 20 cm is the correct answer.

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Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

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