Exam III Practice Problems (liquids and pressures)

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Homework Help Overview

The discussion revolves around practice problems related to liquids and pressures, specifically focusing on concepts such as buoyancy, tension in fluids, and the application of Bernoulli's equation. Participants are preparing for an upcoming exam and are seeking clarification on various physics problems.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on a block submerged in a fluid, including tension, buoyancy, and weight. Questions arise about the application of the Bernoulli equation and the relevance of free body diagrams. Some participants express confusion over terminology and the use of equations related to fluid dynamics.

Discussion Status

There is an ongoing exchange of ideas with some participants providing guidance on identifying forces and applying principles like Archimedes' principle. Others are exploring different interpretations of the problems and questioning their understanding of the concepts involved.

Contextual Notes

Participants mention constraints such as the lack of a textbook and reliance on trial and error due to limited resources. There are also references to specific practice questions from an online resource, indicating that the discussion is grounded in specific exam preparation.

  • #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
 
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  • #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 because 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 because 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 because 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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