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November 01, 2025, 04:30:30 pm
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Author Topic: Quick question  (Read 3461 times)  Share 

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HarveyD

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Quick question
« on: November 07, 2011, 02:02:14 am »
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just wondering for this
why is area enclosed the portion of the bigger circle
would the intersection between the two be the smaller one?
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dc302

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Re: Quick question
« Reply #1 on: November 07, 2011, 02:17:48 am »
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I don't really understand your question. Also, what is L? The question is cut off.
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HarveyD

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Re: Quick question
« Reply #2 on: November 07, 2011, 02:21:54 am »
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L is y = 1/root3 x

how do you find the shaded area?
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dc302

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Re: Quick question
« Reply #3 on: November 07, 2011, 02:31:28 am »
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The shaded area is like a 'piece' of a 'donut' right?

Find the area of the large circle, then subtract the area of the smaller circle. Then find the proportion of the angle of your 'piece'. As in, do you have half the donut? 1/4? 1/9? Etc. Then multiply the area you found before by this proportion.

edit: so I guess, if my assumption was right, then the area of the bigger circle minus the smaller is:

4pi - pi = 3pi

And the proportion of the whole thing is 1/12 (since the angle between the two boundaries is pi/6, and one revolution is 2pi. Then the area required is 3pi*1/12 = pi/4. Is that right?
« Last Edit: November 07, 2011, 02:34:01 am by dc302 »
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Laxzal

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Re: Quick question
« Reply #4 on: November 07, 2011, 10:51:40 am »
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Yes it is pi/4

I hope this explains it to you. WARNING: I suck at paint and I am not the greatest person to explain this kind of stuff, but I shall try!
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HarveyD

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Re: Quick question
« Reply #5 on: November 08, 2011, 09:34:35 pm »
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thanks for the help :)

got another question
with the question below
couldnt C be true as well?
If F = 7 , a = 0
so constant velocity?
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vea

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Re: Quick question
« Reply #6 on: November 08, 2011, 09:43:38 pm »
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The keyword is "moves" with constant velocity. If F=7, it has constant velocity so a=0 however it is at rest.
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Zebra

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Re: Quick question
« Reply #7 on: November 09, 2011, 01:34:43 pm »
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what's the answer?
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Zebra

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Re: Quick question
« Reply #8 on: November 09, 2011, 01:37:27 pm »
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can someone please do this question ?

3 kg : F - T -3g x 1/7 = 3a
2kg  : T - 2g x 1/7 =2 a

??? what next?
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vea

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Re: Quick question
« Reply #9 on: November 09, 2011, 02:04:41 pm »
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Quote from: Zebra on November 09, 2011, 01:37:27 pm
can someone please do this question ?

3 kg : F - T -3g x 1/7 = 3a
2kg  : T - 2g x 1/7 =2 a

??? what next?

Something looks wrong with the equation of motion for the 3kg object, you don't normally have friction and weight force in the same direction. I think the weight force should be in the j direction and you also need a normal reaction force if there is friction. Do you have the full question?
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Zebra

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Re: Quick question
« Reply #10 on: November 09, 2011, 02:07:43 pm »
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no. but i got it. i worked it out! its a?
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HarveyD

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Re: Quick question
« Reply #11 on: November 09, 2011, 02:09:02 pm »
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yep
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vea

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Re: Quick question
« Reply #12 on: November 09, 2011, 02:09:14 pm »
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Quote from: Zebra on November 09, 2011, 02:07:43 pm
no. but i got it. i worked it out! its a?

Nvm, I assumed F was friction. Still not too sure what you're asking but at least you've got the answer. :)
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HarveyD

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Re: Quick question
« Reply #13 on: November 09, 2011, 03:38:33 pm »
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just wondering
how would you guys do this question?
Is it wrong to let
t^2 + 1 = 7t - 5
then
2t = t + 6
then solve from there
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Natters

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Re: Quick question
« Reply #14 on: November 09, 2011, 05:20:24 pm »
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anyone know what the deal is with the abs. value signs in the logs?
wtf is that shit?
sometimes they just make what's in there positive and other times they leave it and i can't work it out
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