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SVETLANKA909090 [29]

2 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

1 answer:

bija089 [108]2 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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2 years ago

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

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Answer:

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Explanation:

From the question we are told that

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The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

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3 years ago

A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.

ValentinkaMS [17]

A) $2.03 m/s^2$

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

$mg sin \theta - \mu_k R = ma$ (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

$\theta=22.5^{\circ}$

$\mu_k = 0.19$ is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

$R-mg cos \theta =0$ (2)

From (2) we find

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

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B) 5.94 m/s

We can solve this part by using the suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

$a=2.03 m/s^2$

s = 8.70 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s$

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3 years ago