R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
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Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

Replacing the diameter the area results:

Therefore the the stress results:

Answer:
t = 25.10 sec
Explanation:
we know that Avrami equation

here Y is percentage of completion of reaction = 50%
t is duration of reaction = 146 sec
so,


taking natural log on both side
ln(0.5) = -k(306.6)

for 86 % completion




t = 25.10 sec
Answer:
The power developed by engine is 167.55 KW
Explanation:
Given that

Mean effective pressure = 6.4 bar
Speed = 2000 rpm
We know that power is the work done per second.
So

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.
P=167.55 KW
So the power developed by engine is 167.55 KW
Answer:
a. 6 seconds
b. 180 feet
Explanation:
Images attached to show working.
a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.
b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.