Exam III Practice Problems (liquids and pressures)

AI Thread Summary
The discussion revolves around solving practice problems related to liquids and pressures for an upcoming exam. Key concepts include the application of the Bernoulli equation, buoyant forces, and the importance of free body diagrams in analyzing forces acting on blocks. Participants clarify misconceptions about equations and emphasize the need to understand underlying principles rather than just plugging in numbers. The conversation also touches on specific problems, such as finding tension and calculating buoyancy, while participants share their approaches and seek guidance on their calculations. Overall, the thread highlights collaborative problem-solving and the importance of conceptual understanding in physics.
  • #51
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 15. I understand it but for some reason it is a 2 star so i just wana make sure i got it for the right reason.

So i knew it was between B and C since it starts from Zero, and velocity must be positive since it said in the problem it is traveling whith a (+) velocity of 11.1 m /s
Not sure I understand your reasoning, but you probably got the correct answer. Explain your thinking when you say "i knew it was between B and C since it starts from Zero?" What starts from zero? (Surely not the velocity.)
 
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  • #52
Doc Al said:
Not sure I understand your reasoning, but you probably got the correct answer. Explain your thinking when you say "i knew it was between B and C since it starts from Zero?" What starts from zero? (Surely not the velocity.)
the time started from zero
 
  • #53
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp04 Question 19 and 21. Same question word for word yet 19.B and 21 is C. Am i missing something?>
They didn't do a good job in wording the question. I suspect that for questions 20 and 21 you are to assume that the plug has been removed so that fluid can flow through the pipe. (Otherwise 20 makes no sense and 21 would be the same as 19, as you realize.)
 
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  • #54
Alt+F4 said:
the time started from zero
That's certainly true, but give your complete reasoning for the answer you think is correct if you'd like me to confirm your thinking.
 
  • #55
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp00 Question 10.

I assumed it would b Lead > Iron

Since (higher Density)(9.8)(Same Volume) > ( Lower Density)(9.8)( Same volume)

but then they turn out to be equal so i think the density to what we plug in is the water density instead of Iron or Lead since the objects are floating in water. Is this correcT?
 
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  • #56
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp00 Question 10.

I assumed it would b Lead > Iron

Since (higher Density)(9.8)(Same Volume) > ( Lower Density)(9.8)( Same volume)

but then they turn out to be equal so i think the density to what we plug in is the water density instead of Iron or Lead since the objects are floating in water. Is this correcT?
Yes. The buoyant force equals the weight of the displaced fluid, so it's the density of water that is used.
 
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  • #57
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 question 10

So what i did was

pgy = pgy

(1000)(9.8)(.3) = (.6)(x)(9.8)
Density of 500 but answer is 666


Edit:


Okay i think this is it

(1000)(9.8)(.7-.3) = (X)(9.8)(.6)
X = 666.67

Is this the correct way
 
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  • #58
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 11. I know i am doing this right

So convert 90 cycles per 1 minute into cycles per second which is 1.5 cycles per sec

T= 2pi/ omega

1.5 = 2pie/ omega
omega = 4.1887

it is a pendulum so

omega^2 = G/L

(4.18879)^2 = 9.8/L

L = .55meters



I only get the answer when i do 60/90 which is 0.66 cycels per sec which makes no sense
 
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  • #59
Alt+F4 said:
Okay i think this is it

(1000)(9.8)(.7-.3) = (X)(9.8)(.6)
X = 666.67

Is this the correct way
That's right. The key here is that the pressure at the interface must be the same on both sides. So the .7-.3= .4m column of water must balance the .6m column of other fluid.
 
  • #60
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 11. I know i am doing this right

So convert 90 cycles per 1 minute into cycles per second which is 1.5 cycles per sec

T= 2pi/ omega

1.5 = 2pie/ omega
omega = 4.1887

it is a pendulum so

omega^2 = G/L

(4.18879)^2 = 9.8/L

L = .55meters



I only get the answer when i do 60/90 which is 0.66 cycels per sec which makes no sense
You are confusing frequency (cycles per second) with period (seconds). To convert frequency to omega (angular frequency, measured in radians per second), multiply by the number of radians per cycle (which you should know).
 
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  • #61
Doc Al said:
You are confusing frequency (cycles per second) with period (seconds). To convert frequency to omega (angular frequency, measured in radians per second), multiply by the number of radians per cycle (which you should know).
I see. Thanks
 
  • #62
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 19


so for that one

P1 = P2 + pgy + .5 pg V^2
P1atm + .5 (9.8)(1000)(2.8^2) + 1000*9.8*2= P2 + (1000)(9.8)(2) + (.5)(1000)(9.8)(2^2)

It says point C is atm pressure so i am using it.

I got 1.88*10^4
Anser is 1.8*10^4 so i wana make sure this is the way it is done

Edit: well i guess u can't do what i did becuase in 21 it asked what is diff of height between B and C, i assumed it was 2
 
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  • #63
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 24-23. This should be it for a while since ima just do all the exam again


so for this problem i was thinking( i guess similar to the Ballastic one)

pgvdi - (mass of Cylinder + mass of Block) * 9.8 - tension

(1000)(9.8)( .005+ (.2 * .3*.3)) = X+ ( i don't know the mass of block) * 9.8 - 24.5

The .2*.3*.3 is the area of the cube,
 
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  • #64
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp02 Question 12. If the elevaotor was Acc downward would the answer have been less than 90 cycles since ur moving downward therefore experience a sligher heavier g
 
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  • #65
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp02 Question 12. If the elevaotor was Acc downward would the answer have been less than 90 cycles since ur moving downward therefore experience a sligher heavier g

Watch out.. aceclerating downward and moving downward are not equivalent. (I am assuming that by ''accelerating downward'' you mean that a_y is negative). You can be moving upward and have a_y negative like you can be moving downward and have a_y positive.

Another thing that confuses me in your post: you consider that it is accelerating downward and yet you talk about the object being slightly ''heavier''?? Don't you mean the opposite?

In any case, if a_y is positive, then the angular frequency will be larger than if the acceleration is zero which means that there will be more oscillations per second.
 
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  • #66
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 19


so for that one

P1 = P2 + pgy + .5 pg V^2
P1atm + .5 (9.8)(1000)(2.8^2) + 1000*9.8*2= P2 + (1000)(9.8)(2) + (.5)(1000)(9.8)(2^2)

It says point C is atm pressure so i am using it.

I got 1.88*10^4
Anser is 1.8*10^4 so i wana make sure this is the way it is done

Edit: well i guess u can't do what i did becuase in 21 it asked what is diff of height between B and C, i assumed it was 2
For this problem, compare a point at the top of the funnel to point B. Note that the top of the funnel is at atmospheric pressure.
 
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  • #67
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 24-23. This should be it for a while since ima just do all the exam again


so for this problem i was thinking( i guess similar to the Ballastic one)

pgvdi - (mass of Cylinder + mass of Block) * 9.8 - tension

(1000)(9.8)( .005+ (.2 * .3*.3)) = X+ ( i don't know the mass of block) * 9.8 - 24.5

The .2*.3*.3 is the area of the cube,
Consider the forces acting on the cube: buoyant force, weight, tension from the string. Since the cube is in equilibrium, those forces must add to zero. Solve for the weight to find the mass.
 
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  • #68
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 19


so for that one

P1 = P2 + pgy + .5 pg V^2
P1atm + .5 (9.8)(1000)(2.8^2) + 1000*9.8*2= P2 + (1000)(9.8)(2) + (.5)(1000)(9.8)(2^2)

It says point C is atm pressure so i am using it.

I got 1.88*10^4
Anser is 1.8*10^4 so i wana make sure this is the way it is done

Edit: well i guess u can't do what i did becuase in 21 it asked what is diff of height between B and C, i assumed it was 2
so i want to make sure i did it right. I don't get why it was marked a 3 star.


P1 + (.5)(1000)(.5^2) + (1000)(9.8)(2) = p2 + (.5)(1000)(2^2)

Pb- Patm = 1.77*10^4 ~ 1.8*10^4
 
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  • #69
Doc Al said:
Consider the forces acting on the cube: buoyant force, weight, tension from the string. Since the cube is in equilibrium, those forces must add to zero. Solve for the weight to find the mass.
okay that was the first thing i tried but it didn't work

Fb-mg-Tension = 0

(1000)(9.8)(.2) - m(9.8) - 24.5 = 0

m = 197.5
 
  • #70
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp01 Question 18

what i did was

pgy = .5p v^2

(1000)(9.8)(.03) = .5(1000)(V^2)

V = .77


Ans: .75

So why am i off by .2
 
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  • #71
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 9. Why is it sin. Time starts from 0 so it should be cos
 
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  • #72
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 9. Why is it sin. Time starts from 0 so it should be cos
The *position* as a function of time is 3 cos (wt). The velocity is the derivative of the position with respect to time so -3 w sin(wt) (I used the chain rule and the fact that the derivative of cos is - sin)
 
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  • #73
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 13. THis should be it, I have done all the exams

Okay so I found Angular Velocity

W^2 = K/M
K = 72 from the above question
72/8 = W^2
W = 3

T= 2pi/3
t= 2.07.
 
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  • #74
nrqed said:
The *position* as a function of time is 3 cos (wt). The velocity is the derivative of the position with respect to time so -3 w sin(wt) (I used the chain rule and the fact that the derivative of cos is - sin)
hmm that was never mentioned in class, and not in lecture notes. He also specified we don't need to know calc, but i get what you are saying. Thanks
 
  • #75
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 13. THis should be it, I have done all the exams

Okay so I found Angular Velocity

W^2 = K/M
K = 72 from the above question
72/8 = W^2
W = 3

T= 2pi/3
t= 2.07.

I bet the answer is half of that? Notice that the block will be in contact with the spring for only half the period (from equilibirum position to the farthest to the right and back to equilibirum)
 
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  • #76
nrqed said:
I bet the answer is half of that? Notice that the block will be in contact with the spring for only half the period (from equilibirum position to the farthest to the right and back to equilibirum)
ohhh ya. Thanks
 
  • #77
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 19.


I found the pressure in Denver which is (1.29)(1.8)(1000)(9.8) = 22755.6

22755.6/1.01*10^5 = .22
Ans:.77
 
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  • #78
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 20. It says Low tide the thing is floats freely on the surface

so what i did was

(1000)(9.8)(X) = (2)(9.8)

X = .002

So .002 is the volume displaced
the Volume of the Entire Cube is .02

.002/.02 = .1
which is not the answer
 
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  • #79
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 20. It says Low tide the thing is floats freely on the surface

so what i did was

(1000)(9.8)(X) = (2)(9.8)

X = .002

So .002 is the volume displaced
the Volume of the Entire Cube is .02
ok but its mass is 2 kg so you could simply use mass over density = 0.02 m^3
.002/.02 = .1
which is not the answer
? you need to find the buoyant force an dthe force of gravity.
Then use -Tension +Buoyant force - mg =0
 
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  • #80
nrqed said:
ok but its mass is 2 kg so you could simply use mass over density = 0.02 m^3

? you need to find the buoyant force an dthe force of gravity.
Then use -Tension +Buoyant force - mg =0
why would there be tension? it is floating freely
 
  • #81
Alt+F4 said:
why would there be tension? it is floating freely
Sorry ! I looked at question 19! Sorry
 
  • #82
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 20. It says Low tide the thing is floats freely on the surface

so what i did was

(1000)(9.8)(X) = (2)(9.8)

X = .002

So .002 is the volume displaced
the Volume of the Entire Cube is .02

.002/.02 = .1
which is not the answer
That's thefraction immersed. They want the fraction above the water line
 
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  • #83
nrqed said:
That's thefraction immersed. They want the fraction above the water line
o yaaaa, i think ima take a break for a while. I am losing it after i did 16 exams. Thanks
 
  • #84
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp01 Question 18

what i did was

pgy = .5p v^2

(1000)(9.8)(.03) = .5(1000)(V^2)

V = .77


Ans: .75

So why am i off by .2
only thing left is just this problem
 
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  • #85
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa03 Question 25

for this one i am just wondering why .1 was used, i would have used .4

(1000)(9.8)(.5) = (1000)(9.8)(.4 instead of .1) + .5 (1000)(V^2)
 
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  • #86
The left-hand side of the equation is describing the condition of water
at the TOP (why at top? because we know the Pressure = 1 atm there!)
The right-hand side describes the condition of the water somewhere else ...
Why would you want to describe water at height h = .4 m?
Where is the water that you are asked about?
Where is the water that has noticeable speed?
 
  • #87
http://online.physics.uiuc.edu/cgi/c...ice/exam3/fa05 Question 24-23. This should be it for a while since ima just do all the exam again


so for this problem i was thinking( i guess similar to the Ballastic one)

pgvdi - (mass of Cylinder + mass of Block) * 9.8 - tension

(1000)(9.8)( .005+ (.2 * .3*.3)) = X+ ( i don't know the mass of block) * 9.8 - 24.5

The .2*.3*.3 is the area of the cube,

Doc Al said:
Consider the forces acting on the cube: buoyant force, weight, tension from the string. Since the cube is in equilibrium, those forces must add to zero. Solve for the weight to find the mass.



Okay so Fbyonat - Tenstion - mg = 0

(1000)(9.8)(.2)(.3)(.3) - 24.5 = m(9.8)

M = 15.5

Edit: That is the mass of Plastic Cube

Now How can i get that small cylinder

(1000)(9.8)(.005) - 24.5 - mg = 0

M=2.5

Ans:7.5
Edit:

This is what i did and got the answer

is this true

(1000)(9.8)((.2*.3*.3)+.005)) = (mass of cube which i got + x) * 9.8

(1000)(9.8)(.023) = (15.5+X)*9.8
X=7.5
 
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  • #88
This link isn't working ...

If the .2x.3x.3 [m^3] plastic block is pulled down by Tension = 24.5 [N] ,
then you use your first equation to solve for the mass of plastic = 15.5 [kg]
(presuming that speed = 0)

If the small .005 [m^3] cylinder is being pulled upward by Tension,
then the Tension 24.5 [N] enters as a POSitive z-component Force.

Are you DRAWING the Force vectors on your diagrams?
 
  • #89
lightgrav said:
This link isn't working ...

If the .2x.3x.3 [m^3] plastic block is pulled down by Tension = 24.5 [N] ,
then you use your first equation to solve for the mass of plastic = 15.5 [kg]
(presuming that speed = 0)

If the small .005 [m^3] cylinder is being pulled upward by Tension,
then the Tension 24.5 [N] enters as a POSitive z-component Force.

Are you DRAWING the Force vectors on your diagrams?
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05

O ya, i guess i need to draw out. I kept subtracting tension Thanks
 
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  • #90
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 14. So i just said K = 2, W = 2 M = 2 and found an imaginary Etotal.

then i found the kinetic energy of .01 and when i divide Etot / 4 i get the kinetic energy that my equation gave. What am i doing wrong
 
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  • #91
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp99 19, absolutely makes no sense. The steel marble will take room in the cup right? thus making the intial level higher unless the water overflows but they never stated that
 
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  • #92
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 14. So i just said K = 2, W = 2 M = 2 and found an imaginary Etotal.

then i found the kinetic energy of .01 and when i divide Etot / 4 i get the kinetic energy that my equation gave. What am i doing wrong
How did you find the KE at 0.01m? Hint: Consider the change in potential energy as the position changes, then use that to find the KE.
 
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  • #93
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp01 Question 18

what i did was

pgy = .5p v^2

(1000)(9.8)(.03) = .5(1000)(V^2)

V = .77


Ans: .75

So why am i off by .2



last question. Any hints?
 
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  • #94
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp99 19, absolutely makes no sense. The steel marble will take room in the cup right? thus making the intial level higher unless the water overflows but they never stated that
Sure the marble takes up room in the cup, but which raises the water level more: having the marble imbedded in the ice, or having the marble tossed in the water? (Hint: Which situation displaces more water?)
 
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  • #95
Alt+F4 said:
last question. Any hints?
You are only off by 0.02 m/s. Your method looks OK to me.
 
  • #96
Doc Al said:
Sure the marble takes up room in the cup, but which raises the water level more: having the marble imbedded in the ice, or having the marble tossed in the water? (Hint: Which situation displaces more water?)
okay so question 18, i understand that it will be the same as inital level, but then 19 there is an object in there that will take room so it will push the water up. I still don't see why the water would be lower? What happened to the water that was at the intial level?

Okay maybe this will clear it, when there saying slightly, how slight are we talking about?
 
  • #97
Alt+F4 said:
okay so question 18, i understand that it will be the same as inital level,
Why is that? What determines the volume of displaced fluid?

but then 19 there is an object in there that will take room so it will push the water up.
It's not just the fact that "it takes room" that's the key. How about this: Imagine a bathtub of water and a big toy boat floating in it. If I take a bowling ball, which will raise the water level in the bathtub more: (1) dropping the ball in the water, or (2) putting the ball on the boat so that the whole thing floats?

The key is: How much water is displaced?
 
  • #98
Thanks everyone for your help, I just aced another exam. I'll be back in a week for finals
 

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