Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
Answer:
Power required to overcome aerodynamic drag is 50.971 KW
Explanation:
For explanation see the picture attached
Answer:
a) P = 86720 N
b) L = 131.2983 mm
Explanation:
σ = 271 MPa = 271*10⁶ Pa
E = 119 GPa = 119*10⁹ Pa
A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²
a) P = ?
We can apply the equation
σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N
b) L₀ = 131 mm = 0.131 m
We can get ΔL applying the following formula (Hooke's Law):
ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)
⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm
Finally we obtain
L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Answer:
diesel fuel is pumped at high pressure to the injectors which are responsible for entering the fuel into the combustion chamber,
when the piston is at the top the pressure is so high that it explodes the fuel (diesel) that results in a generation of mechanical power
