Final Exam Questions

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Okay, I am back for one more time. Last thread of the semster. Thanks for everyone's help for the other 3.


Question 6

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06 [Broken]


How would you go about finding this problem

I know torque = FrSinTheta
 
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Answers and Replies

  • #2
berkeman
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Draw the diagram showing the force on the valve stem due to gravity. What is the angle between the force on the valve stem and the radius from the axle to the valve stem? Do you see why the answer to question 5 also involves this angle?
 
  • #3
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If you know the equation for torque (a vector quantity), then all you have to do is find the magnitude.
 
  • #4
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http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 2, I dont get why it is true

V ave for A = -20 m/s
V ave For B = 10 m/s
V ave For C = 40 m/s

(10+-20+40)/(3) does not equal 0
 
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  • #5
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http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 8

Change in Y is 1

So what i was thinking of doing is

Actually i have no idea

I think the formula i should be using is V^2 = V0^2 + 2a(change in X)

A = 9.8
V0 = 12sin65
so i found V to be 11.74

Then i just used

V = V0 + at
which gave me .084 sec

Any help, thanks
 
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  • #6
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O yes, These types of problems, Can somone just explain how to exactly do it, there are like 20 more on the other exams. I never really got how. Thanks

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 21
 
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  • #7
Doc Al
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Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 2, I dont get why it is true

V ave for A = -20 m/s
V ave For B = 10 m/s
V ave For C = 40 m/s

(10+-20+40)/(3) does not equal 0
Your logic is faulty. Think about this: during period A you earned $100/hour; during period B you earned $1/hour. Are you sure that your average earning rate is (101)/2 = $50.50/hour? What if period A lasted 1 hour but period B lasted 10 years?

The right way to find average velocity is displacement/time.
 
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  • #8
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Doc Al said:
Your logic is faulty. Think about this: during period A you earned $100/hour; during period B you earned $1/hour. Are you sure that your average earning rate is (101)/2 = $50.50/hour? What if period A lasted 1 hour but period B lasted 10 years?

The right way to find average velocity is displacement/time.
I want to thank you for all your help, I aced all the exams, just this final and i am done. Thanks
 
  • #9
Doc Al
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Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 8

Change in Y is 1

So what i was thinking of doing is

Actually i have no idea

I think the formula i should be using is V^2 = V0^2 + 2a(change in X)

A = 9.8
V0 = 12sin65
so i found V to be 11.74

Then i just used

V = V0 + at
which gave me .084 sec

Any help, thanks
Hint: What's the horizontal velocity and acceleration?
 
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  • #10
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Doc Al said:
Hint: What's the horizontal velocity and acceleration?
Horizontal Velocity = 12 Cos 65 = 5.0714

Acc = -9.8

What is confusin me is the Distance factor The Hoop is 3 meters what do i do with that
 
  • #11
Doc Al
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Alt+F4 said:
O yes, These types of problems, Can somone just explain how to exactly do it, there are like 20 more on the other exams. I never really got how. Thanks

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 21
Here's one way to look at it. To get across as fast as possible, just swim straight across (with respect to the water, which will drag you downstream, but who cares?). How long does it take the water to drag the swimmer the given distance downstream? Since he swims across the river in that same time, what must be his swimming speed (with respect to the water)?
 
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  • #12
Doc Al
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Alt+F4 said:
Horizontal Velocity = 12 Cos 65 = 5.0714
Right (in m/s).

Acc = -9.8
Wrong! (What's the horizontal acceleration?)

What is confusin me is the Distance factor The Hoop is 3 meters what do i do with that
What you do with it is ignore it. You don't need it. Focus on the horizontal component of the motion. You have the speed and the distance.
 
  • #13
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Doc Al said:
Here's one way to look at it. To get across as fast as possible, just swim straight across (with respect to the water, which will drag you downstream, but who cares?). How long does it take the water to drag the swimmer the given distance downstream? Since he swims across the river in that same time, what must be his swimming speed (with respect to the water)?
am i making it harder by assuming Vectors and Just figuring it out there cause the triangle is 45 45 90, i can find the angle then from there i can find the speed
 
  • #14
Doc Al
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Alt+F4 said:
am i making it harder by assuming Vectors and Just figuring it out there cause the triangle is 45 45 90, i can find the angle then from there i can find the speed
You can certainly use vectors, but that's more work than needed. I don't see a 45-45-90 triangle.
 
  • #15
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http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 9

So i got 1.97 Sec for 8

so ithinkin of using X = X0 + V0T + .5 a T^2

But that doesnt work
 
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  • #16
Doc Al
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Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 9

So i got 1.97 Sec for 8
OK.

so ithinkin of using X = X0 + V0T + .5 a T^2

But that doesnt work
Why not? It works for me. (Be sure you are using vertical components and proper signs.)
 
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  • #17
306
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Alt+F4 said:
O yes, These types of problems, Can somone just explain how to exactly do it, there are like 20 more on the other exams. I never really got how. Thanks

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 21


Okay so let's see if i got what you are saying

Vave = X/T

1.8 = 55/ T

T = 30.555 sec

V ave = 75 / 30.5555

V ave = 2.5

Is this the correct way?
 
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  • #18
Doc Al
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Alt+F4 said:
Okay so let's see if i got what you are saying

Vave = X/T

1.8 = 55/ T

T = 30.555 sec

V ave = 75 / 30.5555

V ave = 2.5

Is this the correct way?
Looks good to me. The person's speed with respect to the water is 2.5 m/s. (If you wanted to find the person's speed with respect to the shore, you'd have to add the cross-stream (2.5 m/s) and down-stream (1.8 m/s) components--using vector addition--to find his total speed. But you weren't asked that.)
 
  • #19
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okay question 5 and 6

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06 [Broken]


Was question 5 basically you know it or you don't type of thing?

And free body digram would be

Stem Valve Pointing down = mg = .245

T = F r sin theta
 
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  • #20
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Doc Al said:
Looks good to me. The person's speed with respect to the water is 2.5 m/s. (If you wanted to find the person's speed with respect to the shore, you'd have to add the cross-stream (2.5 m/s) and down-stream (1.8 m/s) components--using vector addition--to find his total speed. But you weren't asked that.)


So like (1.8^2) + (2.5^2) = X^2
 
  • #21
Doc Al
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Alt+F4 said:
okay question 5 and 6

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06 [Broken]


Was question 5 basically you know it or you don't type of thing?

And free body digram would be

Stem Valve Pointing down = mg = .245

T = F r sin theta
Careful with your [itex]\sin\theta[/itex]. In the equation for torque, [itex]\theta[/itex] is the angle between the radius vector (r) and the force vector (in this case, mg). That's not the same as the angle [itex]\theta[/itex] shown in the diagram (but they are related, of course).

So what's your answer?
 
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  • #22
Doc Al
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Alt+F4 said:
So like (1.8^2) + (2.5^2) = X^2
Exactly...
 
  • #23
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Doc Al said:
Careful with your [itex]\sin\theta[/itex]. In the equation for torque, [itex]\theta[/itex] is the angle between the radius vector (r) and the force vector (in this case, mg). That's not the same as the angle [itex]\theta[/itex] shown in the diagram (but they are related, of course).

So what's your answer?


Okay so For 5 it is 90 degree so 90 - 24 = 66 and (.025)(9.8)(.3) *sin 66

So where exactly is 66 degrees? is it the one that is drawn? Thanks
 
  • #24
Doc Al
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Alt+F4 said:
Okay so For 5 it is 90 degree
That's the answer to #5: the torque will be minimum (equal to zero!) when theta = 90 degrees.

so 90 - 24 = 66 and (.025)(9.8)(.3) *sin 66
That's the answer for #6. When the angle shown is 24; the angle between force and radius is 90 - 24 = 66 degrees.

So where exactly is 66 degrees? is it the one that is drawn?
66 degrees will be the angle between the force and the radius; or, in other words, between the vertical axis and the radius vector (since the force, gravity, acts vertically). (The diagram shows the angle between the horizontal axis and the radius vector.)
 
  • #25
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http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06 [Broken]

Question 20

Impulese = F ave Change in time = Change in Momentum

Fave ( .07) = ( .125)(Vf ) = (.125)(7)


How do i find Vf

I know it bounces back 2 m

so what i did was

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

V = 6.26 but that doesnt work


Edit: i guess i didnt read the part about it going downward so

( M )( Vf) + (M) (Vi)

Which does work
 
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