List, in ascending order, the cutoff frequencies for the first ten modes of a rectangular waveguide, normalized to the cutoff fr
1 answer:
Answer:
Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22
Explanation:
Assuming a = 2b
Attached below is the required steps to the solution
The cutoff frequencies for the first ten modes of a rectangular waveguide listed in ascending order is :
Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22
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Answer:
the torque capacity is 30316.369 lb-in
Explanation:
Given data
OD = 9 in
ID = 7 in
coefficient of friction = 0.2
maximum pressure = 1.5 in-kip = 1500 lb
To find out
the torque capacity using the uniform-pressure assumption.
Solution
We know the the torque formula for uniform pressure theory is
torque = 2/3 × × coefficient of friction × maximum pressure ( R³ - r³ ) .....................................1
here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in
now put all these value R, r, coefficient of friction and maximum pressure in equation 1 and we will get here torque
torque = 2/3 × × 0.2 × 1500 ( 4.5³ - 3.5³ )
so the torque = 30316.369 lb-in
Answer:
So the fraction of proeutectoid cementite is 44.3%
Explanation:
Given that
cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.
We know that
Fraction of proeutectoid cementite phase gievn as
Now by putting the values
So the fraction of proeutectoid cementite is 44.3%