The answer is 2nd Step because the first step is to define the problem and third is to define your goals
Answer: 5.36×10-3kg/h
Where 10-3 is 10 exponential 3 or 10 raised to the power of -3.
Explanation:using the formula
M =JAt = -DAt×Dc/Dx
Where D is change in the respective variables. Insulting the values we get,
=5.1 × 10-8 × 0.13 × 3600 × 2.9 × 0.31 / 4×10-3.
=5.36×10-3kg/h
Answer:
Technician B
Explanation:
The brakes can lockup due to the following reasons
1) Overheating break systems
2) Use of wrong brake fluid
3) Broken or damaged drum brake backing plates, rotors, or calipers
4) A defective ABS part, or a defective parking mechanism or proportioning valve
5) Brake wheel cylinders, worn off
6) Misaligned power brake booster component
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.
