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.
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hi, i got an exam soon and i need help with a couple of problems

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 7.


How would i find the tension i know that i must use Bernoulli equation but what would i be solving for
 
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What are the forces acting on the block?

Can you draw a free body diagram of the block?

The Bernoulli equation is for fluid flow so I'm not sure why (or how) you would use it here.
 
dav2008 said:
What are the forces acting on the block?

Can you draw a free body diagram of the block?

The Bernoulli equation is for fluid flow so I'm not sure why (or how) you would use it here.
ooo well all i have been doing for the past 4 weeks is using that eqaution for HW and stuff so i thought it must be it. Anywyas, You have the Water pushing up, Wood Pushing down and String pulling down
 
Right. You have the buoyant force acting upwards, the weight of the block acting downwards and tension acting downwards.

Can you see how to solve for tension now?
 
Edit:Never mind
 
You are given the volume of the block and its density. Using those facts you can find its mass.

As far as the forces, like you said earlier there are only 3 forces: Tension down, weight down, and buoyancy up.

Recall Archimedes' principle that the buoyancy force is equal to the weight of the displaced fluid.
 
thanks, I'll be back with more in 5 min. :)
 
dav2008 said:
You are given the volume of the block and its density. Using those facts you can find its mass.

As far as the forces, like you said earlier there are only 3 forces: Tension down, weight down, and buoyancy up.

Recall Archimedes' principle that the buoyancy force is equal to the weight of the displaced fluid.
ya i got it, there are too many stuff that are same like V= Velocity but then it is Volume in that equation. Crazy
 
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 15. ANy trick to finding the quick answer to these. Thanks
 
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  • #10
Well given that it's a multiple choice you can just see which one of those three equations fits your given information.

You're given that at t=0 s v=11.1 m/s.
 
  • #11
Thanks alot, i need to ace this exam so sorry for all the questions.

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 16

Can i use one of these equations for 15 to just plug in 1.5 and get the V which i can get Kinetic energy from
 
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  • #12
Well try it and see if you get the right answer.
 
  • #13
dav2008 said:
Well try it and see if you get the right answer.
:( nope well i was thinking of using X(t) = A cost(omega t) but what would A be hmmmm
 
  • #14
Im stupid. I got it
 
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  • #15
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 19. I am not getting why Pa > Pb. Thinking of it B is a more tighter area so there will be more pressue on it
 
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  • #16
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04 Question 7. SO what i did was

Force of Friction - Force of compression of spring
UkMg - -kx = 0
(X)(20*9.8) = 270 * .2

What am i doing wrong.
 
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  • #17
Question 25 http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04

Okay so i got the answer to 24 which was 600. FOR 25 i am going to use this formula h = (M + m)/ (Density* A)

D= M/v
600 = M/ (2*.15) or is it (2*.25)
little m is just 20*3 = 60
Density on bottom is 1000
A = .25*2
plug that it and i am not getting the answer
 
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  • #18
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 10. hmmmmmmm Well wat i did was

(1000)(9.8)(.3) = (X)(9.8)(.6)
Which gave me 500 which is wrong.
 
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  • #19
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 19. I think this should be it for a while, Thanks for everyones help.

So what i did was

Patm = P1 + .5(1000)(2^2) + (1000*9.8*2)
Which gives me

Patm - PB = 2.2*10^4 (not the answer)

Answer: 1.8*10^4
 
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  • #20
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04 Question 7. SO what i did was

Force of Friction - Force of compression of spring
UkMg - -kx = 0
(X)(20*9.8) = 270 * .2

What am i doing wrong.
I drew out free body diagram and i still can't get it
 
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  • #21
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04 Question 7. SO what i did was

Force of Friction - Force of compression of spring
UkMg - -kx = 0
(X)(20*9.8) = 270 * .2

What am i doing wrong.
For that one


Would it be

Force of friction - (The Total Work of spring) = 0

so

Mu mg - .5KX^2 = 0

but i can't get the answer
 
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  • #22
You can't subtract work from force. It's a meaningless expression

You need a statement of conservation of energy. What energy does the block have when it is pulled .2m, and where does that energy go?
 
  • #23
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 10. hmmmmmmm Well wat i did was

(1000)(9.8)(.3) = (X)(9.8)(.6)
Which gave me 500 which is wrong.
Think about what you are doing. What you said in your expression is that the pressure due to the column of substance b is equal to the pressure of the half column of substance A on the left side. Why did you assume that is true?

Think about which pressures have to be equal.
 
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  • #24
dav2008 said:
You can't subtract work from force. It's a meaningless expression

You need a statement of conservation of energy. What energy does the block have when it is pulled .2m, and where does that energy go?
okay so Potential Energy is .5KX^2 = 5.4 J

Energy then is converted to Kinetic which is also 5.4 J
 
  • #25
You're right that's converted to kinetic but you forgot about friction.

Compare the two states that the block is in:

State 1: all potential energy equal to .5*k*x2

State 2: No potential energy (x=0), no kinetic energy (they told you it's not moving), so the energy must have gone into friction.

Do you know how to find work done by a force over a distance?
 
  • #26
Honestly I think you need to think a bit more about the concepts behind the problems instead of just trying to find an equation to blindly plug numbers into and then posting on here when that equation doesn't work.
 
  • #27
dav2008 said:
You're right that's converted to kinetic but you forgot about friction.

Compare the two states that the block is in:

State 1: all potential energy equal to .5*k*x2

State 2: No potential energy (x=0), no kinetic energy (they told you it's not moving), so the energy must have gone into friction.

Do you know how to find work done by a force over a distance?
okay so are u saying that 5.4J = the energy of friction

So what then do i use to get an asnwer of .14

casue i did 5.4 = X(20*9.8)
 
  • #28
dav2008 said:
Honestly I think you need to think a bit more about the concepts behind the problems instead of just trying to find an equation to blindly plug numbers into and then posting on here when that equation doesn't work.
i would if we had a book, but the lectures as u can see on the site are basics so i just gota do it by trial and error that is why there are so many exams given
 
  • #29
Look at what you're doing.

x(20)(9.8) gives you the force of friction.

That force acts over a certain distance.

Now you want to find the work done by that force.

Do you think you can find it given a force and the distance over which it acts?
 
  • #30
dav2008 said:
Look at what you're doing.

x(20)(9.8) gives you the force of friction.

That force acts over a certain distance.

Now you want to find the work done by that force.

Do you think you can find it given a force and the distance over which it acts?
oooooooooooooooooooo Thanks so much, i forgot totally about W = F*D
 
  • #31
Alt+F4 said:
Question 25 http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04

Okay so i got the answer to 24 which was 600. FOR 25 i am going to use this formula h = (M + m)/ (Density* A)

D= M/v
600 = M/ (2*.15) or is it (2*.25)
little m is just 20*3 = 60
Density on bottom is 1000
A = .25*2
plug that it and i am not getting the answer





Forget what i did upthere

this is what i did

the buyoant force has to balance the weight

Fb = m*9.8
(1000)(9.8)(X) = (300+60)(9.8)

X=.36

.15-.36 = .21 m which is not answer
 
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  • #32
Alt+F4 said:
Forget what i did upthere

this is what i did

the buyoant force has to balance the weight

Fb = m*9.8
(1000)(9.8)(X) = (300+60)(9.8)

X=.36

.15-.36 = .21 m which is not answer
Two problems:

(1) Why do you think the mass of the wood is 300 kg?

(2) The "X" in your equation stands for volume displaced, not depth.
 
  • #33
Doc Al said:
Two problems:

(1) Why do you think the mass of the wood is 300 kg?

(2) The "X" in your equation stands for volume displaced, not depth.
cause i got a density of 600

So what i did was

D= M/V

600 = M/ (2*.25) which gave me 300
 
  • #34
Doc Al said:
Two problems:

(1) Why do you think the mass of the wood is 300 kg?

(2) The "X" in your equation stands for volume displaced, not depth.
well i was thinking since it is already at .15, whatever the number i will get will have the .15 added to it
 
  • #35
Alt+F4 said:
cause i got a density of 600

So what i did was

D= M/V

600 = M/ (2*.25) which gave me 300
Volume requires all three dimesions. You multiplied length by height, but what about width? (If you include units in your calculations, you'll be less likely to make this kind of error.)
 
  • #36
Alt+F4 said:
well i was thinking since it is already at .15, whatever the number i will get will have the .15 added to it
Don't confuse length (or depth) with volume. Length is measured in meters; volume in cubic meters--very different.
 
  • #37
okay so mass is 450 of log, Children are 60 KG

Total = 510 Kg
so...

Boyancy force = Total weight of people + log

(1000)(9.8)(x) = (450+60)(9.8)

ummm i am lost
 
  • #38
Alt+F4 said:
okay so mass is 450 of log, Children are 60 KG

Total = 510 Kg
so...

Boyancy force = Total weight of people + log

(1000)(9.8)(x) = (450+60)(9.8)

ummm i am lost
So far, so good. Realize that x = volume of the displaced water. Volume = Length*Width*Depth (length and width are given).
 
  • #39
Doc Al said:
So far, so good. Realize that x = volume of the displaced water. Volume = Length*Width*Depth (length and width are given).
wow thanks, all this and its a 2 star. Geee
 
  • #40
Sorry back again, http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04 Question 24. I guess i did it wrong and got the right answer or this is the way to do it

So all i did was

(1000)(9.8)(.15) = (9.8)(2*.25)^2(X)

Density i got was 600 which is the answer
 
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  • #41
Alt+F4 said:
Sorry back again, http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04 Question 24. I guess i did it wrong and got the right answer or this is the way to do it

So all i did was

(1000)(9.8)(.15) = (9.8)(2*.25)^2(X)

Density i got was 600 which is the answer
Not sure what that (2*.25)^2 is supposed to be. (But it gives the right answer! :rolleyes: )

Use the same equation as before, with volume = L*W*D:

(1000)(9.8)(L*W)(.15) = X (9.8) (L*W)(.25)
 
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  • #42
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp02 Question 17.

Thanks for your help

I think this should be it for today :)

so i was thinking of using Bernoulli Equation but i guess you would have 2 Unknows since u don't know the volume of air

so i guess you would have to use the Buyoant force equation

so what i did was

(9.8)(1.27)(687) = 8550.402


P.S how can u remb all this stuff. Are you a Prof or a teacher
 
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  • #43
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp02 Question 17.

Thanks for your help

I think this should be it for today :)

so i was thinking of using Bernoulli Equation but i guess you would have 2 Unknows since u don't know the volume of air

so i guess you would have to use the Buyoant force equation

so what i did was

(9.8)(1.27)(687) = 8550.402
This problem has nothing to do with Bernoulli, so you are correct in applying the Buoyant force equation.

What you found is the total mass of the balloon + cargo + hot air; now find just the mass of the hot air by subtracting out the given mass of the balloon + cargo. Then use that to find the density of the hot air. (You can assume that the given volume is the volume of the hot air.)

P.S how can u remb all this stuff. Are you a Prof or a teacher
I can't give away all my secrets. :wink:
 
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  • #44
ooo thanks alot, Night. I'll be back tommarow though :biggrin:
 
  • #45
I lied

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp04 Question 10.

ummm Please tell me this is Bernoulli

What confuses me about these is what do i do with all these masses?
 
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  • #46
Alt+F4 said:
I lied

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp04 Question 10.

ummm Please tell me this is Bernoulli

What confuses me about these is what do i do with all these masses?
I'm afraid I can't tell you that it's a Bernoulli problem. It's a buoyant force problem, just like the others. To raise the ship the buoyant force must be at least enough to equal the weight of the masses: ship + ballast + air.

Note: Bernoulli problems typically involve figuring out changes in pressure or speed along a moving fluid, like water flowing through a pipe of varying diameter.
 
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  • #47
Doc Al said:
I'm afraid I can't tell you that it's a Bernoulli problem. It's a buoyant force problem, just like the others. To raise the ship the buoyant force must be at least enough to equal the weight of the masses: ship + ballast + air.

Note: Bernoulli problems typically involve figuring out changes in pressure or speed along a moving fluid, like water flowing through a pipe of varying diameter.
so what i did was


(1027)(9.8)(X) = (50,000+100)*9.8

How Does air play a role since i did not use the density of air
Anyways i get 48.7, when i divide that by 2 since there are 2 Ballast i get 24.4 which is answer
 
  • #48
Alt+F4 said:
so what i did was


(1027)(9.8)(X) = (50,000+100)*9.8

How Does air play a role since i did not use the density of air
Anyways i get 48.7, when i divide that by 2 since there are 2 Ballast i get 24.4 which is answer
The density of air allows you to compute the mass of the air, which in this case is too small to worry about. A more accurate equation would be:

(1027)(9.8)(X) = (50,000+100 + (1.25*X))*9.8

But that 1.25*X term can be neglected compared to 1027*X. (Note that we are also ignoring the volume of the ship--we assume that it's tiny compared to the ballast tanks. Although it looks big in the drawing.)
 
  • #49
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
 
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  • #50
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?>
 
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