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A refrigerator has a cooling load of 50 kW. It has a COP of 2. It is run by a heat engine which consumes 50 kW of heat to supply
1 answer:
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Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = × Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = ·
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = ×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
Answer:
(a) Yes
(b) 102.8 ft
Explanation:
(a)First let convert mile per hour to feet per second
30 mph = 30 * 5280 / 3600 = 44 ft/s
The time it takes for this driver to decelerate comfortably to 0 speed is
t = v / a = 44 / 10 = 4.4 (s)
given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.
(b) At this rate the distance covered by the driver is
Since the intersection is only 60 feet wide, the dilemma zone must be
162.8 - 60 = 102.8 ft