Answer:
W = 1884J
Explanation:
This question is incomplete. The original question was:
<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.
</em>
<em>
How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.</em>
The amount of work done by the motor is given by:
Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.
By using kinematics:
But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:
=> 
Now we can calculate the final velocity:
Finally, we calculate the total work:
Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":
W = 1884J
A positive or direct relationship is one in which the two variables (we will generally call them x and y) move together, that is, they either increase or decrease together. In a negative or indirect relationship, the two variables move in opposite directions, that is, as one increases, the other descremases
Answer:
We show added energy to a system as +Q or -W
Explanation:
The first law of thermodynamics states that, in an isolated system, energy can neither be created nor be destroyed;
Energy is added to the internal energy of a system as either work energy or heat energy as follows;
ΔU = Q - W
Therefore, when energy is added as heat energy to a system, we show the energy as positive Q (+Q), when energy is added to the system in the form of work, we show the energy as minus W (-W).
Answer:
714.285s
Explanation:
use relative velocity
8-4.5 = 3.5m/s
x = 2500m
2500/3.5 = 714.285s = 700s (with sig figs)
