Answer:
Explanation:
Since the roundabout is rotating with uniform velocity ,
input power = frictional power
frictional power = 2.5 kW
frictional torque x angular velocity = 2.5 kW
frictional torque x .47 = 2.5 kW
frictional torque = 2.5 / .47 kN .m
= 5.32 kN . m
= 5 kN.m
b )
When power is switched off , it will decelerate because of frictional torque .
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer:
Explanation:
The acceleration of a circular motion is given by
where 
is the angular velocity and 
is the radius.
Angular velocity is related to the period, T, by
Substitute into the previous formula.
This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.
Answer: A; True
Explanation: Momentum is known to be a vector quality, and thus has been proven by modern scientists and resulting in this answer being true.
Hope this helps <3
Stay safe, stay warm
-Carrie
Ps. it would mean a lot if you marked brainliest (=
