6. Discussion: 4. What is the relationship type in the plot between (Fexp) versus (p QVJ)? Why? 5. Is there a difference between the slope from plot and the slope of the value (1-Cos)? Why?
6. Discussion: 4. What is the relationship type in the plot between (Fexp) versus (p QVJ)? Why? 5. Is there a difference between the slope from plot and the slope of the value (1-Cos)? Why?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Transcribed Image Text:6. Discussion:
4. What is the relationship type in the plot between (Fexp) versus (p Q VJ)?
Why?
5. Is there a difference between the slope from plot and the slope of the
value (1- Cos)? Why?
Transcribed Image Text:Calculation
time = 28, Volume = 5, mass: 70, h = 5. __
VOL
5
1=0.17 m²/sec
(V₁ = = 0.17
time
28
12.56
A = I ² = = ~(4) = 12.56 m² / V5= Vn -√2gh
= 0.013-√2016)
=83.64
FJ=9Q VJ (1-cose)
=(1/(0-17) (83.64) (1-C0545)
= 4.265
90-45, time = 32, Volume = 5, mass=140
/V₁=
Q = Vol = 5= 0.15 m²/sie
tim 32
..15
12.56
A=Id=-(6)=12.56² / V5= Vn -√2gh
=0.011-√√2x 140.50
= 118.3
FJ=A@v₂ (1-cosys)
=(1)(0.15) (118.3) (0.31
= 5.323
J
0=45
= 0.013 mhu
h: 50mm
= 0.011 m/s.c
Calculation
D 0=9 time=19, mass: 1oo, ho 50mm, Volume 5
Vol 50.26 m²/see
time 19
/ Vni
V₁ = @=
0.26
12.56
A= d² = (16) = 12.56 m² / V5= V₁-√2gh
=0.02-√2x100 50
= 100
FJ=PQV₂ (1-C059)
=(1) (0-26) (100) (1-0)
= 26
90-9:
time = 23, mass = 70, h = 50mm, Volum. 5
VOL
5
0.21
/V₁==
= 0.21 m²/91.
1256
23
A=Id': (6) = 12.56 m²
VS-Vu-Jegh
= 0.016-√2+70×5
FJ= POV, (1-cosq:)
= 83.64
=(1) (0.21) 183.64)(1)
= 17.56
= 0.02 m/sec
= 0.016 m/s.
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