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sergey [27]
3 years ago
7

A 3 cm tall object is placed 2 cm to the left of a converging lens that has a focal length with a magnitude of 4 cm. A diverging

lens with a focal length of magnitude 8 cm is placed 10 cm to the right of the first lens.
What is the magnification of the final image produced by these two lenses? Make sure you say whether it is bigger/smaller and upright/inverted.
Physics
1 answer:
kramer3 years ago
8 0

Answer:

the mangnafication of the final is smaller anbsgs sgs shs sgs vd

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A linear accelerator uses alternating electric fields to accelerate electrons to close to the speed of light. A small number of
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Answer:i

9E13 ELECTRONS

Explanation:

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So no of electrons is now q/e

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To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

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KE = \frac{1}{2} mv^2

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m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

m*V_i = m_1V_1+m_2V_2

Here,

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m_1=m_2 = Mass each part

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V_1 = Final velocity particle 1

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KE_i=\frac{1}{2}mv^2

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By conservation of the moment then,

m*V_i = m_1V_1+m_2V_2

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(8)*5 = 4*V_1+4*V_2

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In the final point the cinematic energy of EACH particle would be given by

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V_2 = 10-V_1

200J=\frac{1}{2}4*(V_1^2+V_2^2)

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: V_1 = 10m/s

Then replacing in (1) we have that

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