Hey guys,
I am currently studying for my physics 2 exam and want to post up the exam and my answers and see whether you guys agree or disagree with them, since I cannot find solutions to it.
Anyway q1:
a) D
b) C
c) If bernoulli's equation states:
_{1}+\rho*g*h_{1}=p_{2}+\dfrac{1}{2}*\rho*(v^{2})_{2}+\rho*g*h_{2})
Then if I let point 1 be at the surface of the lake (i.e. 60m above C) and point 2 be at C, then this will mean p
1 = p
2, so I can cancel these out:
_{1}+\rho*g*h_{1}=\dfrac{1}{2}*\rho*(v^{2})_{2}+\rho*g*h_{2})
I can set h
2 = 0 => h
1 = 60m =>
_{1}+588,000=\dfrac{1}{2}*\rho*(v^{2})_{2})
By continuity:
so if A
1 >> A
2 => v
2 >> v
1 =>
=>
_{2})
EDIT ERROR (v
2 = v
c =
24.25m/s), v
c = 34.3m/s
By continuity:
=>
^{2}*v_{B}=\pi*(\dfrac{5}{1000})^{2}*34.31)
EDIT ERROR v
B =
0.43m/s, v
B = 0.61m/s
d) Pick point A and point B
bernoulli's equation states:
_{A}+\rho*g*h_{A}=p_{B }+\dfrac{1}{2}*\rho*(v^{2})_{B}+\rho*g*h_{B})
again let:
- v
A = 0
- h
B = 0
- p
A = 1 atm = 101,325 Pa
p
B = 689,139 Pa
p
B(Gauge) = 587,814 Pa
e)
R = 2.694m
3 /s => R = 2.694 L/s
f) Since I am considering the bottom to be at C, I can assume all mechanical energy is kinetic energy.
^{2}}{t} = \dfrac{1}{2}*\rho*R*(v_{C})^{2} = \dfrac{1}{2}*1000*2.694*10^{-3}*34.3^{2} = 1585W)
EDIT: After reviewing my answer to C all my answers matched up to the exam's answers
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