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
(a) 296.6 m
The motion of the stone is the motion of a projectile, thrown with a horizontal speed of
and with an initial vertical velocity of
where we have put a negative sign to indicate that the direction is downward.
The vertical position of the stone at time t is given by
(1)
where
h is the initial height
g = -9.81 m/s^2 is the acceleration due to gravity
The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:
y(6.00 s) = 0
Substituting into eq.(1), we can solve to find the initial height of the stone, h:
(b) 176.6 m
The balloon is moving downward with a constant vertical speed of
So the vertical position of the balloon after a time t is
and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:
(c) 198.2 m
In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.
The horizontal speed of the rock is
So the horizontal distance travelled in t = 6.00 s is
Considering also that the vertical height of the balloon after t=6.00 s is
The distance between the balloon and the rock can be found by using Pythagorean theorem:
(di) 15.0 m/s, -58.8 m/s
For an observer at rest in the basket, the rock is moving horizontally with a velocity of
Instead, the vertical velocity of the rock for an observer at rest in the basket is
Substituting time t=6.00 s, we find
(dii) 15.0 m/s, -78.8 m/s
For an observer at rest on the ground, the rock is still moving horizontally with a velocity of
Instead, the vertical velocity of the rock for an observer on the ground is now given by
Substituting time t=6.00 s, we find
The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
<h3>What is the law of conservation of momentum?</h3>According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of hockey player 1= 91.0-kg
(m₂) is the mass of hockey player 2= 91.0-kg
(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.
(u₂) is the velocity before the collision of hockey player 2=?
a)
Momentum before the collision;
Momentum before the collision = 1237.6 kg m/s².
b)
The velocity of the two hockey players after the collision from the law of conservation of the momentum as:
Momentum before collision = Momentum after the collision
1237.6 kg m/s² = (m₁+m₂)V
1237.6 kg m/s² =(2 ×91.0-kg )V
V=6.8 m/sec.
Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
To learn more about the law of conservation of momentum refer;
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