A horse does 910 J of work in 380 seconds while pulling a wagon. What is the power output of the horse? Round your answer to two
2 answers:
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Answer:
The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.
Explanation:
As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:
Since the frictional force is equal to
, then we have that the acceleration is:
Now, from the definition of acceleration we get:
And, as the final velocity is zero because the box gets to a stop, we have:
(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)
Then, plugging in the given values, we calculate the time:
In words, the time the box takes to stop sliding relative to the belt is 0.139s.
The displacement of the box in this time, is given by the kinematics formula:
Finally, we calculate the displacement:
This means that the box moves 7.29cm backwards relative to the belt.
Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.