A. I believe, lmk if I’m right
Answer: 255
255 turns are required to create 25 ohms of secondary impedance.
Explanation:
Given that,
Number of turns in primary wire N₁ = 900
impedance on Primary wire Z₁ = 400 ohms
Number of turns in Secondary wire N₂ = ?
impedance on Secondary wire Z₂ = 25 ohms
we know that, the relationship between turn and impedance is
Zp / Zs = ( Np / Ns )²
(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²
there fore
Z₁ / Z₂ = ( N₁ / N₂ )²
Now we substitute
( 400 / 25 ) = ( 900 / N₂ )²
400 / 25 = 900² / N₂²
we cross multiple to get our N₂
400 × N₂² = 900² × 25
N₂² = ( 900² × 25 ) / 400
N₂² = ( 810000 × 25 ) / 400
N₂² = 20250000 / 400
N₂² = 50625
N₂ = √50625
N₂ = 225
Therefore 255 turns are required to create 25 ohms of secondary impedance.
Answer: The force exerted on the dough.
Explanation:
The force is responsible for stimulating the stress.
Recall that:
stress= Tensile force/area.
Answer:
modulus of elasticity for the nonporous material is 340.74 GPa
Explanation:
given data
porosity = 303 GPa
modulus of elasticity = 6.0
solution
we get here modulus of elasticity for the nonporous material Eo that is
E = Eo (1 - 1.9P + 0.9P²) ...............1
put here value and we get Eo
303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )
solve it we get
Eo = 340.74 GPa
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