Register to reply

Final Exam Questions

by Alt+F4
Tags: exam, final
Share this thread:
Alt+F4
#73
May10-06, 05:13 PM
P: 306
Quote Quote by Hootenanny
No, I'm afraid it isn't right. The reaction force (mg) is due to the tractor's weight only. The F is the force required to accelerate the total mass at 1.4 m.s-2. Do you follow?

~H
o yes, thanks
Alt+F4
#74
May10-06, 05:57 PM
P: 306
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp03

Question 23

I did Centripital Acceleration * weight

50 * .5 = 25 is this the right way?
Doc Al
#75
May10-06, 06:32 PM
Mentor
Doc Al's Avatar
P: 41,315
Quote Quote by Alt+F4
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp03

Question 23

I did Centripital Acceleration * weight

50 * .5 = 25 is this the right way?
That's fine. (I assume you meant to write mass, not weight.)
Alt+F4
#76
May10-06, 07:20 PM
P: 306
http://online.physics.uiuc.edu/cgi/c...ice/exam2/fa05 Last question, i still dont get how to set it up incase he asks the same question and wants the resultant velocity

so

(1.5*10^5)(8) = (-4*1.5*10^5) + (1.5*10^6*X)

Is this the setup
Doc Al
#77
May10-06, 07:40 PM
Mentor
Doc Al's Avatar
P: 41,315
Quote Quote by Alt+F4
http://online.physics.uiuc.edu/cgi/c...ice/exam2/fa05 Last question, i still dont get how to set it up incase he asks the same question and wants the resultant velocity

so

(1.5*10^5)(8) = (-4*1.5*10^5) + (1.5*10^6*X)

Is this the setup
That's how I'd do it.
Alt+F4
#78
May10-06, 07:54 PM
P: 306
http://online.physics.uiuc.edu/cgi/c...ice/exam2/sp05

Question 6

So what i was thinkin is to do it this way


so work done by friction = W = *.43*9.8*12 = 50.568

Total Kinetic Energy = MGH + .5 MV^2 = (9.8)(9) + .5 (20^2) = 288.2

288.2-50.568 = 237.632

Edit: well it works if i dont take in effect the PE of the hill but why not?
Doc Al
#79
May10-06, 08:11 PM
Mentor
Doc Al's Avatar
P: 41,315
Quote Quote by Alt+F4
http://online.physics.uiuc.edu/cgi/c...ice/exam2/sp05

Question 6

So what i was thinkin is to do it this way


so work done by friction = W = *.43*9.8*12 = 50.568

Total Kinetic Energy = MGH + .5 MV^2 = (9.8)(9) + .5 (20^2) = 288.2

288.2-50.568 = 237.632

Edit: well it works if i dont take in effect the PE of the hill but why not?
Why should the 9 m hill matter? Since there's no friction, the KE is the same before and after climbing that hill.
Alt+F4
#80
May10-06, 09:05 PM
P: 306
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
Alt+F4
#81
May10-06, 10:10 PM
P: 306
http://online.physics.uiuc.edu/cgi/c...ice/exam2/fa04

Question 16

So i know that Change in kinetic = Uk g * D

So what i did was (22)(9.8) + (9)(9.8) = X * 9.8 *53

wat am i doing wrong
Hootenanny
#82
May11-06, 03:33 AM
Emeritus
Sci Advisor
PF Gold
Hootenanny's Avatar
P: 9,781
Quote Quote by Alt+F4
http://online.physics.uiuc.edu/cgi/c...ice/exam2/fa04

Question 16

So i know that Change in kinetic = Uk g * D

So what i did was (22)(9.8) + (9)(9.8) = X * 9.8 *53
Your problem is this equation. Why have you added additional GPE [+ (9)(9.8)]?

~H
Alt+F4
#83
May11-06, 11:41 AM
P: 306
http://online.physics.uiuc.edu/cgi/c...ice/exam2/fa04

Question 19

So Wnet = 8.6483
Wnet - Force of friction - Force of gravity
I did get the above questions right so i know the numbers are right


8.6483 = .5 mV^2

V = 2.9773


The 3 is the ending velocity
(-2)(2.9773) - (2)(3) = -11.9546 ( Total Momemtum)

Fave T = P

X (.06) = (11.9546)

X = 200

Is this how it is done
Alt+F4
#84
May11-06, 12:07 PM
P: 306
WHen it says you Double The Frequency does it Actually mean Multiply it by 2 or does it mean Half it?
Hootenanny
#85
May11-06, 12:31 PM
Emeritus
Sci Advisor
PF Gold
Hootenanny's Avatar
P: 9,781
Quote Quote by Alt+F4
WHen it says you Double The Frequency does it Actually mean Multiply it by 2 or does it mean Half it?
In what context? What is the full question?

~H
Hootenanny
#86
May11-06, 12:35 PM
Emeritus
Sci Advisor
PF Gold
Hootenanny's Avatar
P: 9,781
Quote Quote by Alt+F4
http://online.physics.uiuc.edu/cgi/c...ice/exam2/fa04

Question 19

So Wnet = 8.6483
Wnet - Force of friction - Force of gravity
I did get the above questions right so i know the numbers are right


8.6483 = .5 mV^2

V = 2.9773


The 3 is the ending velocity
(-2)(2.9773) - (2)(3) = -11.9546 ( Total Momemtum)

Fave T = P

X (.06) = (11.9546)

X = 200

Is this how it is done
Why have you gone through the trouble of calculating the velocity just before impact when it is given in the question? But yes you working is correct. The change in momentum will be -6m and you were correct to use the impulse relationship.

~H
Alt+F4
#87
May11-06, 12:55 PM
P: 306
Suppose the mass of the string in the above problem is M0. What is the mass of a new string (in terms of M0 ), if the fundamental frequency of the string is doubled compared to the previous value, but the length and tension in the string are the same as before?

Now does this Mean Frequncy * 2 or frequency / 2
Alt+F4
#88
May11-06, 12:56 PM
P: 306
Quote Quote by Alt+F4
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
let me just quote this so it doesnt get lost
Hootenanny
#89
May11-06, 12:58 PM
Emeritus
Sci Advisor
PF Gold
Hootenanny's Avatar
P: 9,781
Quote Quote by Alt+F4
Suppose the mass of the string in the above problem is M0. What is the mass of a new string (in terms of M0 ), if the fundamental frequency of the string is doubled compared to the previous value, but the length and tension in the string are the same as before?

Now does this Mean Frequncy * 2 or frequency / 2
It means the fundemetal fequency is twice what is was before. I.e. If intially the fundemental frequency was 50Hz, the new fundemental frequency would be 100Hz.

~H
Hootenanny
#90
May11-06, 01:00 PM
Emeritus
Sci Advisor
PF Gold
Hootenanny's Avatar
P: 9,781
Quote Quote by Alt+F4
let me just quote this so it doesnt get lost
I answered this above in post #86

~H


Register to reply

Related Discussions
Final Exam Help Introductory Physics Homework 3
Complex numbers (practice exam questions for exam in 2 days) Calculus & Beyond Homework 6
Final exam help! Introductory Physics Homework 2
Questions for Final Exam Introductory Physics Homework 7
Final exam questions: estimators. Introductory Physics Homework 3