A body hanging from a spring balance stretches it by 2cm at the earth,s surface .How much the same body stretch the spring at a place 4800km above the earth,s surface ? Radius of the earth = 6400 km....
plzzz give me solution.....
Answers
Answered by
1
Answer:
In equilibrium, weight of the suspended body = stretching force.
∴ At the earth's surface, mg=k×x
At a height h, mg
′
=k×x
′
g
g
′
=
x
x
′
=
(R
e
+h)
2
R
e
2
=
(6400÷1600)
2
(6400)
2
=(
8000
6400
)
2
=
25
16
x
′
=
25
16
×x=
25
16
×1cm=0.64cm
Explanation:
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2
Answer:
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