Exam III Practice Problems (liquids and pressures)

In summary: Well it is released and the rest of the energy is converted to heat. What was the force of friction then well i have no ideaIn summary, the conversation is about a student preparing for an exam and asking for help with various physics problems. They discuss using the Bernoulli equation, drawing free body diagrams, and solving for tension and forces. They also discuss using multiple choice questions to find the correct equations and using the formula for Archimedes' principle. The conversation also includes multiple questions about specific problems and the process of solving them, including using conservation of energy.
  • #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.
 
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  • #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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  • #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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