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.
The KVA rating of the step down transformer at the given power factor would be 62.5 kVA.
<h3>
What is power factor of a transformer?</h3>
Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA).
PF = working power / apparent power
PF = kW/kVA
kVA = kW/PF
kVA = 50 kW/0.8
kVA = 62.5 kVA
Thus, the KVA rating of the step down transformer at the given power factor would be 62.5 kVA.
Learn more about power factor here: brainly.com/question/7956945
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Answer:
The Current will decrease by a factor of 2
Explanation:
Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.
Now, as the question says, the load is reduced to half its original value, we can write:


Since, P2 = P1/2,

Dividing equations (1) and (2), we get,
P1 / (P1/2) = I1/ I2

Hence, it is proved that the current in the transmission line will decrease by a factor of 2 when load is reduced to half.
Answer:
Both model building codes and NFPA 220 can be used to determine the type of construction used in a building.
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