Answer:
ω = 0.467 rad/s
Explanation:
given,
tangential force exerted by the person = 37.7 N
radius of merry-go-round = 2.75 m
mass of merry-go-round = 144 Kg
angle = 33.2°
moment of inertia
I = 544.5 kg.m²
torque = force x radius
τ = 37.7 x 2.75
τ = 103.675 N.m
angular acceleration
α = 0.190 rad/s²
now ,
distance =
d = 0.579 rad
we know,
using equation of rotational motion
t = 2.46 s
angular speed
ω = α x t
ω = 0.19 x 2.46
ω = 0.467 rad/s
The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
a = g = -9.8
downward (acceleration due to gravity).
By using Suvat equation:
v = u + at
where: v is the velocity at time t
u = 40.0 m/s is the initial velocity
a = g = -9.8 is the acceleration
To find the time t at which the velocity is v = 20.7 m/s
Therefore,
The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
The complete question is:
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?
To learn more about uniformly accelerated motion refer to:
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