Incomplete question.The complete question is here
Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.
Answer:
Torque=0.51 Btu
Explanation:
Given Data
Power=225 hp
Revolutions =3000 rpm
To find
T( torque )=?
Solution
As
As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.
So
Answer:
after shock
creating a system for the conservation of the energy of the basketball ball and creating a system for the tennis ball only, the conservation of energy should be applied to each system independently
Explanation:
When the two balls fall they acquire the same speed since they are accelerated by the same force, their weight and the acceleration of the acceleration of gravity. When reaching the floor the mechanical energy of the system is conserved.
Upon reaching the floor, the first ball (basketball) collides with the floor, this process is very fast, at the end of the process the basketball comes out with a velicad up and collides with the much lighter tennis ball that is still descending .
we assume that the shocks are elastic, when solving the momentary and kinetic energy findings, we find the velocities after each shock
In this clash the tennis ball acquires a high kinetic speed with an upward direction that makes a very high height high. Again this shock is very fast and the tennis ball almost does not move.
Here we must separate the system, creating a system for the conservation of the energy of the basketball ball and another system for the tennis ball only, the conservation of energy should be applied to each system independently
Em₀ =K = 1/2 m v²
= U = m g h
As in the elastic shock the final speed of the tennis ball is approximately 2 vo, we can calculate the maximum height
m g h = 1/2 m (2v₀)²
h = 2 v₀²/g
To reconcile this with the conservation of energy we must calculate the energy for the tennis ball at two points, the first when the crash with the tennis ball ends and at the end point at its maximum height.
Answer:
B. QC > 0; QH < 0
Explanation:
Given that there are two reservoir of energy.
Sign convention for heat and work :
1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.
2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.
From hot reservoir heat is going out that is why it is taken as negative
From cold reservoir heat is coming inside the reservoir that is why it is taken as positive
That is why the answer will be
,