Please explain this solution especially the highlighted part. Why does h^-1/2dh became h1-h2? Expert Answer ⒸStep1a)d₁=200m or 0-20md₂ 200m or 0-20mC=0.80For Rectangular part,t₁ =1₁=2024 125/150worshton.1-8= (021-12 [15-11-2t₁ = 1578-1355For Triangular part,2 =Given.t₂ =Tube-Aven, A₂ = Jy²Plan, A, 30 x 15 = 450m²a sodbArgh 200)) (0-1² savŠ.dh=2021-125[1/2] ²30 ⇒x=30h15 (20) bdb= (122-843).2 [√3-√T-Et₂ = 1807.7495= 1578.135+1807-749T= 3385.88 sBY56.43134 min: The time needed to empty the pool through theseBY56.43134min.tubes is 3385-88 seconds:: T= t₁tt₂= x (01)²2009-03 =1122-847

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Given. d₁=20cm or d₂ 20cm or C = 0·80 For Rectangular port, t₁ = "S" Adh ₂ 21h 1-8 3 4₁ = 2021-125/1 db=2021-125 1-8 =600-[-] t₁ = =1578-1355 For Triangular part, = 30 >x= 10 h h 1-8 t₂ = 15 (2016) bdb ✓ 2(0-2) (7) (0-1)² √19-62 = (1122-842). 2 [√2-√T-E] t₂ = 1807.7495 T= t₁tt₂ = 1578.135+1807.749 0.20m Tube-Aven, A₂= 7z2 0-20m Plana A, 30 x 15 = 450m² a sodb 3- 2 (0-8)(x) (0-1)² √2x9-81xh (1/₂) + 22-847 دالت db

Given. d₁=20cm or d₂ 20cm or C = 0·80 For Rectangular port, t₁ = "S" Adh ₂ 21h 1-8 3 4₁ = 2021-125/1 db=2021-125 1-8 =600-[-] t₁ = =1578-1355 For Triangular part, = 30 >x= 10 h h 1-8 t₂ = 15 (2016) bdb ✓ 2(0-2) (7) (0-1)² √19-62 = (1122-842). 2 [√2-√T-E] t₂ = 1807.7495 T= t₁tt₂ = 1578.135+1807.749 0.20m Tube-Aven, A₂= 7z2 0-20m Plana A, 30 x 15 = 450m² a sodb 3- 2 (0-8)(x) (0-1)² √2x9-81xh (1/₂) + 22-847 دالت db

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Ans : 1 - Fce = 2.74 KN 2 - Fbe = 2.10 KN
A 0.65kg, 50m tape was standardized and supported througnout its whole length and found to be 0.00205m longer at an observed temperature of 31.8°C and a pull of 10KN. This tape was used to measure a 4% grade line which was found to be 662.702m. During measurement, the temperature is 15°C and the tape is suspended under a pull of 20KN. E=200GPa, cross- sectional area of tape is 3mm? and the coefficient of linear expansion is 0.0000116m/°C.
Use any method ans is 697.5lb
Use MDM. a=3.6 b=36.2 c=33.6
P or F value = 36
W2 = 15kN/m a = 2 b = 3 P1 = 75kN
L=8m
10. TAPE MEASUREMENTSa. A line was recorded as 1750.00 meters long. It is measured with a 0.65 kg standard tape30.00m long at 20°C under a 50N pull supported at both ends. During the measurementthe temperature was 5°C and the tape was suspended under a 75N pull. It was also foundthat the tape is 3.0mm too short. The line was inclined on a 3% grade. What is the trueinclined distance? E = 200GPa, A = 2.50mm² and the coefficient of linear expansion is0.0000116 m/°C.b. A line was recorded as 2350.00 meters long. It is measured with a 0.65 kg standardtape 30.00m long at 20°C under a 50N pull supported at both ends. During themeasurement the temperature was 5°C and the tape was suspended under a 75N pull.It was also found that the tape is 3.0mm too short. The line was inclined on a 3% grade.What is the true inclined distance? E = 200GPa, A = 2.50 mm² and the coefficient of linearexpansion is 0.0000116 m/°C.
Only 1 nd 2.
ASAP a=4 b=2
Use w=789
I need hand written solution onlyd=6 in D=12 i z1= 6 f z2 = 12 f Q= 13 cfsttn