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Pressure = lops. 2F₂=F₁ -FR = (Q (O.-V₁ FR=F₁ - PQ (0-v₁) PA) + P Q Determine the magnitude of the force & ter les Icfs pef. that must be to resist the resultant force Q₁=1 cfs Q₂ = 0.75 cfs THE >X yin ↓ TFS Q₁ =a25cfs lied at the tee applied 10

Pressure = lops. 2F₂=F₁ -FR = (Q (O.-V₁ FR=F₁ - PQ (0-v₁) PA) + P Q Determine the magnitude of the force & ter les Icfs pef. that must be to resist the resultant force Q₁=1 cfs Q₂ = 0.75 cfs THE >X yin ↓ TFS Q₁ =a25cfs lied at the tee applied 10

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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I wanted to use the formula that was written in the gray picture. I don't want it to look like the white picture.

O pressure = lops.. Q₁ = 1cfs 4 Q₂ = 0.75 cfs TV F₂ 4in. FR 4.n. 2F₂=F₁ - FR = 1Q (O.-V₂ FR=F₁ - PQ (0-V₁) ↓ TF Q₂ = 25 cfs H+PQ. 하 EP₁A P Determine the magnitude of the force of 12 fr legs that must be applied at the tee to resist the resultant force.
Transcribed Image Text:O pressure = lops.. Q₁ = 1cfs 4 Q₂ = 0.75 cfs TV F₂ 4in. FR 4.n. 2F₂=F₁ - FR = 1Q (O.-V₂ FR=F₁ - PQ (0-V₁) ↓ TF Q₂ = 25 cfs H+PQ. 하 EP₁A P Determine the magnitude of the force of 12 fr legs that must be applied at the tee to resist the resultant force.
Write the net force equation for forces along horizontal x-axis. F. = 0 net,x Fp,1+Fjet, 1-Rx=0 P₁Ac+m₁(V₁0)-Rx=0 P₁Ac+m₁V₁-Rx=0 R₂=P₁A₁+₁V₁=P₁Ac+ (0Q₁) A pQ₁² Ac = P₁Ac+. 2 = ²₁ (70²)+ ( 12 )0, ² π (1² = (10 psi)( (4 in)²) + = 147.87 lb 62.4 o 32.2 lb ft s² X 1 ftª ft³ 1 124 in 4 S ( 1 (4 in)²) X 12³ in ³ 1 ft³ 2
Transcribed Image Text:Write the net force equation for forces along horizontal x-axis. F. = 0 net,x Fp,1+Fjet, 1-Rx=0 P₁Ac+m₁(V₁0)-Rx=0 P₁Ac+m₁V₁-Rx=0 R₂=P₁A₁+₁V₁=P₁Ac+ (0Q₁) A pQ₁² Ac = P₁Ac+. 2 = ²₁ (70²)+ ( 12 )0, ² π (1² = (10 psi)( (4 in)²) + = 147.87 lb 62.4 o 32.2 lb ft s² X 1 ftª ft³ 1 124 in 4 S ( 1 (4 in)²) X 12³ in ³ 1 ft³ 2
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