Answer:
a) 3.242
b) 1291.178 KJ
c) 3.59 cents
Explanation:
a) First all the temperature units should be in Kelvin. -30°C + 273.15 = 243.15K and 45°C + 273.15 = 318.15K. Coefficient of performance for refrigerator is COP = TL/(TH-TL) where TL is Lower Temperature and TH is the Higher Temperature. COP = 243.15/(318.15-243.15) = 3.242
b) COP = Cooling Effect/Work Input. Cooling Effect (or Desired Output) is 4186 kJ.
So Work Input = Cooling Effect/COP = 4186KJ/3.242= 1291.178 KJ
c) 3.6 MJ (3600kJ) is for 10 cents, using ratio for cost: energy used

3.59 cents
Explanation:
time period= 1/frequency.
= 1/50 = 0.02 second.
hope this helps you.
Assume that the deceleration due to braking is a ft/s².
Note that
40 mph = (40/60)*88 = 58.667 ft/s
25 mph = (25/60)*88 = 36.667 ft/s
The final velocity is zero when the car stops, therefore
v² - 2ad = 0, or d = v²/(2a)
where
v = initial speed
a = deceleration
d = stopping distance.
The stopping distance, d₄₀, at 40 mph is
d₄₀ = 58.667²/(2a)
The stopping distance, d₂₅, at 25 mph is
d₂₅ = 36.667²/(2a)
Therefore
d₄₀/d₂₅ = 58.667²/(2a) ÷ 36.667²/(2a)
= (58.667/36.667)²
= 2.56
Answer:
The stopping distance at 40 mph is 2.56 times the stopping distance at 25 mph.
Explanation:
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