Potential energy decreases and kinetic energy increases.
Potential energy is related to the height, since the wagon is going downhill, height decreases and potential energy decreases.
Kinetic energy is related to the speed, since the wagon is speeding up, kinetic energy increases.
Answer:
Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 
to 
is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area
As velocity is in 
and time is in 
so the unit of area is 
Hence, the total distance is 
.
For Blake:
3 boxes at a distance of 10 meters each, each box weighs 20 N
Work done by Blake = 3 * 10m * 20N
= 600 J
Power = 600 J/ 2 min
= 300 J/min
For Sandra:
4 boxes, 15 N each at a distance of 12 meters each.
Work done by Sandra = 4 * 15 N *12m
= 720 J
Power = 720 J/ 4 min
= 180 J/min
Blake does less work than Sandra.
Blake's power is more than Sandra's.
Answer: 1026s, 17.1m
Explanation:
Given
COP of heat pump = 3.15
Mass of air, m = 1500kg
Initial temperature, T1 = 7°C
Final temperature, T2 = 22°C
Power of the heat pump, W = 5kW
The amount of heat needed to increase temperature in the house,
Q = mcΔT
Q = 1500 * 0.718 * (22 - 7)
Q = 1077 * 15
Q = 16155
Rate at which heat is supplied to the house is
Q' = COP * W
Q' = 3.15 * 5
Q' = 15.75
Time required to raise the temperature is
Δt = Q/Q'
Δt = 16155 / 15.75
Δt = 1025.7 s
Δt ~ 1026 s
Δt ~ 17.1 min
