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Oksanka [162]
4 years ago
13

A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the

Physics
1 answer:
Umnica [9.8K]4 years ago
6 0

Answer:

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

H_{max}=\frac{V^2\sin ^2 \theta}{2g}

g = 9.8m/s2

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

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A diverging lens with a focal length of 14 cm is placed 12 cm to the right of a converging lens with a focal length of 21 cm. An
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-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

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0.023 m  right of diverging lens

Explanation:

given data

focal length f2 = 14 cm = -0.14 m

Separation s = 12 cm = 0.12 m

focal length f1 = 21 cm = 0.21 m

distance u1 = 38 cm

to find out

final image be located and Where will the image

solution

we find find  image location i.e v2

so by lens formula v1 is

1/f = 1/u + 1/v     ...............1

v1 = 1/(1/f1 - 1/u1)

v1 = 1/( 1/0.21 - 1/0.38)

v1 = 0.47 m

and

u2 = s - v1

u2 = 0.12 - 0.47

u2 = -0.35

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v2 = 1/(1/f2 - 1/u2)

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v2 = 0.023

so here 0.023 m  right of diverging lens

6 0
3 years ago
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