Answer:
≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
= 
× CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, 
We can as well say:
We can now replace 
in the above equation.
Therefore, = × CρAv²
The A which stands as the area of the jet is given by the formula:
We can now have a new equation after substituting our A into the previous equation as:
= × Cρ 
Substituting our data from above; we have:
= × 
= 
= 110,990N
in N (newton) to KN (kilo-newton) will be:
= 
= 110.990 KN
≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.
Answer:
both statement is correct
Explanation:
Flywheel engine uses to reduce fluctuations.
And
FlexPlate is a metal disk that connects the output from the engine to the input of the torque converter. This will replace the flywheel
so that both statement is correct
Answer:
= 
in thirty seconds.
Explanation:
one thirty second is one part out of 32 equal section . It is used to describe amounts accurately.
can be easily expressed as
Answer:
b). Occurs at the outer surface of the shaft
Explanation:
We know from shear stress and torque relationship, we know that
where, T = torque
J = polar moment of inertia of shaft
τ = torsional shear stress
r = raduis of the shaft
Therefore from the above relation we see that
Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.
When r= 0, then τ = 0
and when r = R , τ is maximum
Thus, torsional shear stress is maximum at the outer surface of the shaft.
