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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
Answer:
<h2>a)Explanation:
If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;
S is the altitude of the ballest
u is the initial velocity
a is the acceleration of the body
t is the time taken to strike the ground
Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s
Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g
Substituting this values into the equation of the motion;
a) An expression for the altitude of the ballast after t seconds is therefore
b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);
This means that the ballast strikes the ground after 6secs
c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.
v = 0 + 32(6)
v = 192ft