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eduard
3 years ago
8

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Physics
1 answer:
Pachacha [2.7K]3 years ago
4 0

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

u -v= d

u=71+26=97\ cm

We need to calculate the focal length

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

put the value into the formula

\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}

\dfrac{1}{f}=-\dfrac{71}{2522}

f=-35.52\ cm

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value in to the formula

\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}

\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}

\dfrac{1}{v}=-\dfrac{124}{2755}

v=-22.21\ cm

We need to calculate the magnification

Using formula of magnification

m=\dfrac{v}{u}

m=\dfrac{22.21}{95}

m=0.233

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

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Answer:

4.33m

Explanation:

Power = work done/ time

work done = power × time =650 × 2 = 1300J

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According to Newton's Third Law of Motion, what happens when two objects of unequal masses collide?
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8 0
2 years ago
Diamond has an index of refraction of 2.419. What is the critical angle for internal reflection inside a diamond that is in air?
SSSSS [86.1K]

Answer:

C) 24.4°

Explanation:

let nd = 2.419 be the index of refraction of diamond and na  = 1.0 be the index of refraction of air  and ∅c be the critical angle.

according to Snell's Law:

sin(∅c) = na/nd

sin(∅c) = (1.0)/(2.419)

      ∅c = 24.4°

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3 years ago
What is the source of the radioactivity?
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4 0
3 years ago
A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
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