A man pushed the box with a force of 20 N. He doubled his force to 40 N. The box did not
1 answer:
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Answer:
Acceleration,
Explanation:
Given that,
Height from a ball falls the ground, h = 17.3 m
It is in contact with the ground for 24.0 ms before stopping.
We need to find the average acceleration the ball during the time it is in contact with the ground.
Firstly, find the velocity when it reached the ground. So,
u = initial velocity=0 m/s
a = acceleration=g
It is in negative direction, u = -18.41 m/s
Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.
So, the average acceleration of the ball during the time it is in contact is .
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
