How much work is done lifting a 12-m chain that is initially coiled on the ground and has a density 2 kg/m so that its top end i
1 answer:
Answer:
W = 627.2 J
Explanation:
Given:
= 2kg/m
length of chain = 12 m
length pulled will be = 8 m
We know
Work done (W) = mgh
where
m = mass of the object
g= acceleration due to gravity
h = displacement
For a small length dy of the chain, the work done can be written as:
dW = (mass of the small length pulled)× g×dy
dW = 2kg/m ×dy×9.8×y
where, y is the distance from the ground level of the end of chain
integerating the above equation
W =
W =
W = 627.2 J
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Answer:
1.3 rev/s
Explanation:
= Moment of inertia when arms and one leg of the skater is out = 3.5 kgm²
= Moment of inertia when arms and legs of the skater are in = 0.80 kgm²
= Angular speed of skater when arms and legs of the skater are in = 5.5 rev/s
= Angular speed of skater when arms and legs of the skater are out = ?
Using conservation of angular momentum
=
(0.80) (5.5) = (3.5)
= 1.3 rev/s