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Wittaler [7]
3 years ago
8

Light rays from an object after passing through a convex lens form an image at the focal point behind the lens (opposite side of

the object). What is the actual position of the object? at the focal point in front of the lens at twice the focal length between the focal point and twice the focal length at a great distance, effectively infinite between the focal point and the lens NextReset
Physics
2 answers:
Colt1911 [192]3 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.


The actual position of the object is between the focal point and twice the focal length and <span>at the focal point in front of the lens. I think it would be A and C. </span>
BaLLatris [955]3 years ago
3 0

Answer:

effectively infinite between the focal point and the lens

Explanation:

Light rays from an object after passing through a convex lens form an image at the focal point behind the lens (opposite side of the object).

Now from lens formula we know that

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

now we know that image forms at focal point

so we will have

d_i = f

now from the equation above

\frac{1}{f} + \frac{1}{d_o} = \frac{1}{f}

so we will have

\frac{1}{d_o} = 0

so position of object is at infinite distance from lens

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Answer:

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Explanation:

When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression

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As we can see this expression is a form of Newton's second law

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Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

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He climbs 2.75-m ladder that is leaning against a vertical wall.

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The angle between the weight of the painter and the displacement is :

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W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J

Hence, the required work done is 1786.17 J.

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