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puteri [66]
2 years ago
6

A slide with an image 4cm×2cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen

1.1 m from the slide. the size of the image on the screen is​
Physics
1 answer:
pav-90 [236]2 years ago
7 0

Answer:

44cm x 22cm

Explanation:

u= 10 cm

v= 1.1 cm

m=v/u= 1.1/10

m=11

hence the size of the image.

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m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
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m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
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v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
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p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

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