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padilas [110]
3 years ago
11

A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig

ht of the image? (note: solve for the image distance first)
Physics
1 answer:
SCORPION-xisa [38]3 years ago
3 0

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : f=20\ cm

Height of object : h=1\ cm

Object distabce from lens : u=-10\ cm

Using lens formula: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}, we get

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}, where v = image distance from the lens.

On solving aboive equation , we get

\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm

Formula of Magnification : m=\dfrac{v}{u}=\dfrac{h'}{h} , where h' is the height of image.

Put value of u, v and h in it , we get

\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm

Hence, the height of the image is 2 cm.

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Answer:

50 degree.

Explanation:

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The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:

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Substitute Ay and Ax into the formula above.

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3 years ago
Read the scenario and solve these two problems.
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Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy K of an object is given by:

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Where:

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V=30 m/s is the speed of the train

Solving the equation:

K=\frac{1}{2}(12000 kg)(30 m/s)^{2}

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P_{i}=0 J is the train's initial potential energy

K_{f}=0 J is the train's final kinetic energy

P_{f}=mgh is the train's final potential energy, where g=9.8 m/s^{2} is the acceleration due gravity and h is the height.

Rewriting the equation with the given values:

5400000 J=(12000 kg)(9.8 m/s^{2})h

Finding h:

h=45.918 m \approx 45.92 m

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