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April 24, 2024, 08:17:13 pm

Author Topic: Predictions for this year's Unit 3 Exam  (Read 9389 times)  Share 

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DisaFear

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Re: Predictions for this year's Unit 3 Exam
« Reply #90 on: June 12, 2011, 06:30:16 pm »
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In electronics, what is an analogue system?



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Re: Predictions for this year's Unit 3 Exam
« Reply #91 on: June 12, 2011, 06:57:01 pm »
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In analogue system the signal can take a continuous range of values, unlike a digital system that takes only a set range of values.
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thatisanote

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Re: Predictions for this year's Unit 3 Exam
« Reply #92 on: June 13, 2011, 06:19:53 pm »
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So can anyone confirm for me:
Banked curve
Fn=Fg/(cos(theta)) ? Or does Fn=Fg*cos(theta)
netF=mgtan(theta)?
netF=Ffriction*cos(theta)+N*sin(theta)
mv^2/r=mgtan(theta) + Ffriction*cos(theta)

I'm sure one of those is wrong...
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adelaide.emily10

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Re: Predictions for this year's Unit 3 Exam
« Reply #93 on: June 13, 2011, 06:30:28 pm »
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So can anyone confirm for me:
Banked curve
Fn=Fg/(cos(theta)) ? Or does Fn=Fg*cos(theta)
netF=mgtan(theta)?
netF=Ffriction*cos(theta)+N*sin(theta)
mv^2/r=mgtan(theta) + Ffriction*cos(theta)

I'm sure one of those is wrong...
i think that is all right except:
Normal force=mg/cos(theta) not mg*cos(theta)

ttn

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Re: Predictions for this year's Unit 3 Exam
« Reply #94 on: June 13, 2011, 06:34:22 pm »
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So can anyone confirm for me:
Banked curve

mv^2/r=mgtan(theta) + Ffriction*cos(theta)

Yeah, that's what my tutor got as well.
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cranberry

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Re: Predictions for this year's Unit 3 Exam
« Reply #95 on: June 13, 2011, 06:38:23 pm »
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is that for friction acting up or down the slope? or does it not matter?
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Bozo

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Re: Predictions for this year's Unit 3 Exam
« Reply #96 on: June 13, 2011, 07:03:04 pm »
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Ok, i need a diagram of like one of those questions they'd ask you about a car. And the forces acting through it. Stationary and moving, because i know that something happens with the friction?

Asx4Life

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Re: Predictions for this year's Unit 3 Exam
« Reply #97 on: June 13, 2011, 07:10:50 pm »
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Is question 9 of insight 2011 wrong? How did they get the clipping at 1v and 7v? Isn't it supposed to be clipping at 1v and 7v instead???

johnbob12

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Re: Predictions for this year's Unit 3 Exam
« Reply #98 on: June 13, 2011, 07:15:19 pm »
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So can anyone confirm for me:
Banked curve

mv^2/r=mgtan(theta) + Ffriction*cos(theta)

Yeah, that's what my tutor got as well.

Shouldn't it be mv^2/r=mgtan(theta)+Ff/cos(theta)

mv^2/r=Nsin(theta)+Ffcos(theta)
Ncos(theta)=mg+Ffsin(theta)
Then solve for N in equation 2 and sub it into equation 1.

« Last Edit: June 13, 2011, 07:17:27 pm by johnbob12 »

adelaide.emily10

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Re: Predictions for this year's Unit 3 Exam
« Reply #99 on: June 13, 2011, 07:16:16 pm »
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i clipped it at 1v and 9v - my teacher said the answer for that question is wrong

Asx4Life

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Re: Predictions for this year's Unit 3 Exam
« Reply #100 on: June 13, 2011, 07:23:07 pm »
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i clipped it at 1v and 9v - my teacher said the answer for that question is wrong

Yeah, I meant 1v and 9v. Also  question 7 of structures and materials is wrong right? I friggin followed their calculations, yet I got 2,250,000 instead of their answer of 500,000,000. this is pissing me off

adelaide.emily10

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Re: Predictions for this year's Unit 3 Exam
« Reply #101 on: June 13, 2011, 08:16:53 pm »
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i clipped it at 1v and 9v - my teacher said the answer for that question is wrong

Yeah, I meant 1v and 9v. Also  question 7 of structures and materials is wrong right? I friggin followed their calculations, yet I got 2,250,000 instead of their answer of 500,000,000. this is pissing me off

yep q7 is wrong - the correct answer is 2.5 x 10^6 or something along those lines

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Re: Predictions for this year's Unit 3 Exam
« Reply #102 on: June 13, 2011, 09:50:31 pm »
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How did you guys derive:

Fc=mgtan(theta)
 
??
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xZero

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Re: Predictions for this year's Unit 3 Exam
« Reply #103 on: June 13, 2011, 10:01:36 pm »
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banked curves, vertical net force = 0 since the object is moving in a horizontal circle. So Ncos(theta)=mg, since centripetal force is Nsin(theta), Fc=Nsin(theta), Fc=mg*sin(theta)/cos(theta). Fc=mgtan(theta)
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