How much work is required to stop a car of mass 1100 kg that is traveling at 27 m/s? 15kJ
1 answer:
Answer:
Work done, W = 401 kJ
Explanation:
It is given that,
Mass of the car, m = 1100 kg
Initial, velocity of the car, u = 27 m/s
Finally it stops, v = 0
According to work energy theorem, the change in kinetic energy is equal to the work done i.e.
W = -400950 J
or
-ve sign shows the direction of force and displacement are in opposite direction.
So, 401 kJ work is required to stop a car. Hence, this is the required solution.
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Answer:
The value of total energy needed per minute for the humidifier = 77.78 KJ
Explanation:
Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)
Specific heat of water = 4190
The heat of vaporization is = 2256
Mass = 0.030 kg
Energy needed to heat water to boiling point =
Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)
Energy () = 10.08 KJ
Energy needed to convert liquid water into vapor at the boiling point
= 0.030 × 2256 = 67.68 KJ
Thus the total energy needed E = +
E = 10.08 + 67.68
E = 77.78 KJ
This is the value of total energy needed per minute for the humidifier.