Efficiency = Power Output / Power Input
Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h
= 222 MJ/h
But 1 hour = 3600seconds
222 MJ/h = 222 MJ/3600s = 0.061667 MW J/s = Watts
Power input = 0.061667 MW = 61 667 W
From Efficiency = Power Output / Power Input
28% = Power Output / 61667
Power Output = 0.28 * 61667
Power Output = 17266.76 W
Power Output = 17 267 W
Rate of heat Rejection = Power input - Power output
= 61667 - 17267 = 44400 W
Rate of heat Rejection = 44 400 W.
C- Copyright.
Answer:
I know I am a very good answerable teacher but I can't answer this question I don't know what
<span>d.rotating counterclockwise and slowing down
This is a matter of understanding the notation and conventions of angular rotations. Positive rotations are counter clockwise and negative rotations are clockwise. An easy way to remember this is the "right hand rule". Make a closed fist with your right hand and have the thumb sticking outwards. If you orient your thumb such that it's pointing in the direction of the positive value along the axis, your fingers will be curled in the positive rotational direction. So in the described scenario, the sphere is rotating in the positive direction (counter clockwise) and decelerating due to the negative angular acceleration. That immediately indicates that options "a", "b", and "e" are wrong since they mention the sphere going clockwise at the beginning. Of the two remaining options "c" and "d", we can discard option "c" since it has the rotation speeding up, and that leaves us with option "d" where the sphere is rotating counter clockwise and slowing down.</span>
Answer:
h = 20 m
Explanation:
given.
height, h = 10 m
Potential energy at 10 m = 50 J
Kinetic energy at 10 m = 50 J
maximum height the ball will reach, H = ?
Total energy of the system
T E = 50 J + 50 J
T E = 100 J
now,
A h = 10 m
P E = m g h
50 = m g x 10
mg = 5 ..............(1)
at the top most Point the only Potential energy will be acting on the body.
now, TE = Potential energy
100 = m g h
5 h = 100
h = 20 m
hence, the maximum height reached by the ball is equal to 20 m.
Two or more velocities add by vector addition
