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When the child jumps onto the merry-go-around the moment of inertia of the system changes. If we consider the child to be point-like mass then its moment of inertia would be:
We get the new moment of inertia by simply adding the child's moment of inertia to the old moment of inertia.
Since there is no force mention we must assume that angular momentum is conserved.
When we plug in all the numbers we get:
We get the new moment of inertia by simply adding the child's moment of inertia to the old moment of inertia.
Since there is no force mention we must assume that angular momentum is conserved.
When we plug in all the numbers we get:
13.06m
Explanation:
Given parameters:
initial velocity u = 0
final velocity v = 16m/s
Unknown;
height of the hill = ?
Solution:
We are going to apply one of the laws of motion to solve this problem;
V² = U² + 2gH
where V is the final velocity
U is the initial velocity
g is the acceleration due to gravity = 9.8m/s²
H is the height
since initial velocity is zero;
V² = 2gH
To find the height;
H =
H = = 13.06m
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