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.
The next step is -748 divide by -11 is 68 m (answer) the pic got cropped sorry
Answer:
option 1
Explanation:
i just used the SOH CAH TOA, and since the given is tan=opposite/adjacent, that should be the answer
The energy stored in a capacitor is given by:

where
U is the energy
C is the capacitance
V is the potential difference
The capacitor in this problem has capacitance

So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
Four of them pass a point every second.