Final Exam Questions

by Alt+F4
Tags: exam, final
P: 306
 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
 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?
Mentor
P: 41,439
 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.)
 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
Mentor
P: 41,439
 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.
 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?
Mentor
P: 41,439
 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.
 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?
 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
Emeritus
PF Gold
P: 9,772
 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

~H
 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
 P: 306 WHen it says you Double The Frequency does it Actually mean Multiply it by 2 or does it mean Half it?
Emeritus
PF Gold
P: 9,772
 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
Emeritus
PF Gold
P: 9,772
 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
 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
P: 306
 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
Emeritus
PF Gold
P: 9,772
 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
Emeritus
PF Gold
P: 9,772
 Quote by Alt+F4 let me just quote this so it doesnt get lost
I answered this above in post #86

~H

 Related Discussions Introductory Physics Homework 3 Calculus & Beyond Homework 6 Introductory Physics Homework 2 Introductory Physics Homework 7 Introductory Physics Homework 3