A relaxed biceps muscle requires a force of 25.0N for an elongation of 3.0 cm; under maximum tension, the same muscle requires a
1 answer:
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Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w = =
= 18.38 rad/sec
Period of oscillation =
= 0.34 sec
Hi there!
We can use the conservation of angular momentum to solve.
Recall the equation for angular momentum:
We can begin by writing out the scenario as a conservation of angular momentum:
= moment of inertia of the merry-go-round (kgm²)
= angular velocity of merry go round (rad/sec)
= final angular velocity of COMBINED objects (rad/sec)
= moment of inertia of boy (kgm²)
= angular velocity of the boy (rad/sec)
The only value not explicitly given is the moment of inertia of the boy.
Since he stands along the edge of the merry go round:
We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:
Plug in the given values:
Now, we must solve for the boy's moment of inertia:
Use the above equation for conservation of momentum: