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March 28, 2024, 10:52:33 pm

Author Topic: VCE Physics Question Thread!  (Read 603199 times)  Share 

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PB

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Re: VCE Physics Question Thread!
« Reply #495 on: July 10, 2014, 01:08:27 pm »
+1
289 W
Yep, I am not sure what you did wrong but your formula was right. Pp=VpIp
we have Vp so we need to find Ip.   Ip= Vp/R =170/100 =1.7

Pp=1.7*170 = 289W

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Rishi97

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Re: VCE Physics Question Thread!
« Reply #496 on: July 10, 2014, 01:37:08 pm »
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Yep, I am not sure what you did wrong but your formula was right. Pp=VpIp
we have Vp so we need to find Ip.   Ip= Vp/R =170/100 =1.7

Pp=1.7*170 = 289W

Thanks heaps for that PB :)
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knightrider

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Re: VCE Physics Question Thread!
« Reply #497 on: July 10, 2014, 02:21:09 pm »
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seeing as you are going from v/t to pos/t graph this time. ANTIdifferentiation is required (area under the graph=displacement).  You can figure out the v-t graph formulas this time as they are linear so it is possible to antidiff.

Thanks for your help but i still dont understand what will be my position points for my position/time grapgh and how do i work them out

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Re: VCE Physics Question Thread!
« Reply #498 on: July 10, 2014, 02:56:14 pm »
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Thanks for your help but i still dont understand what will be my position points for my position/time grapgh and how do i work them out
Ok, I just realised that you don't even need to antidiff any formulas. You can just count the squares - its the same thing...area under the graph and all.
So at t=1 the area under the graph is 1m/s  x 1 s = 1m (seconds cancel out to give you metres - the position!).  t=2   ---->  1m/s x 2s = 2m.  t=3   ---->  3 and a half squares so 3.5m  So on and so forth.
Just find the area under the graph at each second and plot those values on the Pos/time graph. Then connect the dots.  Do you understand how that works?
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knightrider

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Re: VCE Physics Question Thread!
« Reply #499 on: July 10, 2014, 03:40:25 pm »
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Ok, I just realised that you don't even need to antidiff any formulas. You can just count the squares - its the same thing...area under the graph and all.
So at t=1 the area under the graph is 1m/s  x 1 s = 1m (seconds cancel out to give you metres - the position!).  t=2   ---->  1m/s x 2s = 2m.  t=3   ---->  3 and a half squares so 3.5m  So on and so forth.
Just find the area under the graph at each second and plot those values on the Pos/time graph. Then connect the dots.  Do you understand how that works?


yes i do thanks so much :)

PB

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Re: VCE Physics Question Thread!
« Reply #500 on: July 10, 2014, 04:34:38 pm »
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yes i do thanks so much :)

Cool beans :) BTW, I have a physics lecture on next Tuesday  which teaches you how to deal with exactly these kinds of stuff and includes shortcuts for some questions and much much more. Would you be interested in coming?
I have held this lecture before and it seems to have been very beneficial for those who attended! Alll the details can be found in the link in my signature below. Please do check it out! It will be the last time I will be holding it, so don't miss out!
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Rishi97

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Re: VCE Physics Question Thread!
« Reply #501 on: July 10, 2014, 07:17:18 pm »
0
What is the difference between a slip-ring and a split ring communtator?
Which is better and why?
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Re: VCE Physics Question Thread!
« Reply #502 on: July 10, 2014, 08:04:31 pm »
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how would you do these questions

What is the acceleration of the train 10 s after starting?



What is the acceleration of the train 40 s after starting?




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Re: VCE Physics Question Thread!
« Reply #503 on: July 11, 2014, 11:06:00 am »
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The acceleration is given by the gradient, I think. So, if you can find the gradient of the line (generally found by drawing a straight line at the position you want, ie. 10s and for the second question 40s) then you can use the rise over run formula to find the gradient which gives you the acceleration.

eg. the acceleration at 10s I would say is 20/20 = 1m/s^2.

This is described here (scroll down to finding the gradient of a curve): http://www.mathsrevision.net/gcse-maths-revision/algebra/gradients-and-graphs

Does that make sense? Hope it helps,
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Re: VCE Physics Question Thread!
« Reply #504 on: July 11, 2014, 11:56:33 am »
+2
What is the difference between a slip-ring and a split ring communtator?
Which is better and why?

They do completely different things. Slip rings don't change current direction => AC
Split rings do reverse current direction => DC
Hopefully you learn how split rings reverse the current direction
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knightrider

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Re: VCE Physics Question Thread!
« Reply #505 on: July 11, 2014, 01:01:02 pm »
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The acceleration is given by the gradient, I think. So, if you can find the gradient of the line (generally found by drawing a straight line at the position you want, ie. 10s and for the second question 40s) then you can use the rise over run formula to find the gradient which gives you the acceleration.

eg. the acceleration at 10s I would say is 20/20 = 1m/s^2.

This is described here (scroll down to finding the gradient of a curve): http://www.mathsrevision.net/gcse-maths-revision/algebra/gradients-and-graphs

Does that make sense? Hope it helps,Stewart



i still dont get how you got 20/20 = 1m/s^2. because it is a curve so you cant do rise on run and if you did that it would be 20/10=2m/s^2  the answers say something else to both the answers  can anyone help

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Re: VCE Physics Question Thread!
« Reply #506 on: July 11, 2014, 01:45:00 pm »
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Hi there,

What is the answer?

Check these links for finding the gradient of a non-linear graph, you have to draw a line and find the gradient of that line as the other link explains. This is the acceleration.
http://cstl.syr.edu/fipse/graphb/unit8/unit8a.html
http://www.columbia.edu/itc/sipa/math/slope_nonlinear.html
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Re: VCE Physics Question Thread!
« Reply #507 on: July 11, 2014, 02:18:17 pm »
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as I mentioned before knightrider. This type of question will never be tested by VCAA - there will be too much variation in the answers. My advice is to not spend too much time on these types of questions and focus on other more important stuff.
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Re: VCE Physics Question Thread!
« Reply #508 on: July 11, 2014, 04:13:45 pm »
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A sphere of mass 12kg is allowed to roll down a plane inclined at 30 degrees to the horizontal. There is a constant frictional force of 20N acting on the sphere.
WHat is the acceleration?
Using the a=gsin(angle) formula, I am getting the answer of 5m/s/s but the answer says 3.3m/s/s

What am I doing wrong? Why doesn't the formula work?

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Re: VCE Physics Question Thread!
« Reply #509 on: July 11, 2014, 04:49:57 pm »
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You need to take away the acceleration in the opposite direction caused by the frictional force.
a = F/m = 20/12 = 1.7m/s/s
So the 'net' acceleration would be 5-1.7=3.3m/s/s