if the accepted thickness of aluminum foil is 1.5x10^5 is your answer percise, accurate or both. explain your answer
1 answer:
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Answer:
19.95 J
Explanation:
The center of mass of the ladder is initially at a height of:
The center of mass of the ladder ends at a height of:
=L/2
So, the work done is equal to the change in potential energy which is:
W = PE =
now
therefore
W = [mgL/2]×[1 - sin(theta)]
W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]
solving this we get
W = 19.95 J
Answer:
Explanation:
Let the extension in the spring be x .
restoring force = weight of block
kx = mg
x =
= 23.84 cm
b )
When the elevator is going upwards
Restoring force = mg + ma
k x₁ = 10.9 ( 9.8 + 1.89 )
x₁ = 28.44 cm
( y coordinate will be - ( 28.44 - 23.84 ) = - 4.6 cm )
c ) When the cable snaps , both elevator and block undergo free fall . In this case apparent g = 0
Since the spring is stretched by 28.44 cm , a restoring force continues to act on the block which is equal to
.2844 x 448
= 127.41 N
So a net acceleration a will act on the block
a = 127.41 / 10.9
= 11.68 m / s²
The block will undergo SHM with amplitude equal to 28.44 cm .