Answer:
3.3619 Nm
54.27472 rad
182.46618 J
86.88 W
Explanation:
= Initial angular momentum = 7.2 kgm²/s
= Final angular momentum = 0.14 kgm²/s
I = Moment of inertia = 0.142 kgm²
t = Time taken
Average torque is given by

Magnitude of the average torque acting on the flywheel is 3.3619 Nm
Angular speed is given by

Angular acceleration is given by

From the equation of rotational motion

The angle the flywheel turns is 54.27472 rad
Work done is given by

Work done on the wheel is 182.46618 J
Power is given by

The magnitude of the average power done on the flywheel is 86.88 W
Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2
Answer:
Explanation:
It's equal to change of momentum in contact
Explanation:
Q = mc∆T
= (0.34 kg)(94 J/kg-°C)(25°C)
= 799 J
8a2-10ab+15b+10 Explaintion: