g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],
1 answer:
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4.75cm * 5.22cm is 24.795cm²
but let's do this with m² instead:
0.0475m * 0.0522m = 0.0024795
now we can compare it with the 49780 much easier.
devide. that's the <u>roughly</u> 20,000,000fold of the plot, but it's still squared, so let's take the root
so the representative fraction is 1:4,480.
the exact value is in the screens
have a nice day:)
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.
