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March 30, 2024, 12:23:09 am

Author Topic: VCE Physics Question Thread!  (Read 603602 times)  Share 

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Ancora_Imparo

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Re: VCE Physics Question Thread!
« Reply #480 on: June 24, 2014, 12:03:31 pm »
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Whoops, just got the last line wrong.

if

Zealous, in the post above, also got the same, but just used , which is why I believe the answer is slightly different.
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knightrider

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Re: VCE Physics Question Thread!
« Reply #481 on: June 28, 2014, 09:37:29 pm »
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How would you do the following question

Homer

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Re: VCE Physics Question Thread!
« Reply #482 on: June 28, 2014, 09:49:58 pm »
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a)50+30=80km
b)50-30=20km north
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Stew_822

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Re: VCE Physics Question Thread!
« Reply #483 on: June 28, 2014, 09:55:06 pm »
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Distance is a scaler value so you simply add the two together. Since displacement is a vector, you have to do vector addition, so if we define north as positive, south would be in the opposite direction and hence negative.
So 50 + (-30) = +20km and since we have said positive is north, the full answer is 20km north.
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knightrider

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Re: VCE Physics Question Thread!
« Reply #484 on: June 28, 2014, 10:40:20 pm »
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Thankyou for your help guys

but in my book it says displacement=final position-initial position

in this case it would be 30-50=-20 which would be south                 

Can anyone explain how this works

Yacoubb

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Re: VCE Physics Question Thread!
« Reply #485 on: June 29, 2014, 01:19:36 am »
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Thankyou for your help guys

but in my book it says displacement=final position-initial position

in this case it would be 30-50=-20 which would be south                 

Can anyone explain how this works

So total distance would be 50km + 30km = 80km (no direction needed as this is a scalar quantity).

Now, let's make south positive, and thus north negative.

-50km + 30km = -20km
So, we know that the displacement is 20km NORTH.

Let's try making south negative, and thus north positive.

-30km + 50km = 20km
So, displacement = 20km north (given that this value is positive, denoting north).

Hope this helps (:

Stew_822

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Re: VCE Physics Question Thread!
« Reply #486 on: June 29, 2014, 10:02:01 am »
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It still works using that definition, so again we will say north is positive. Let's also say his initial position is 0. I'll do a little diagram for you:

Code: [Select]
- - - - - >
     <- - -
So he moves 50km right (north) then 30km south. His initial position is 0, then +50, then he moves back 30km so his final position is +20km.

Using that formula, displacement = final position - initial = +20 - 0 = +20km and since positive then its north.

An easier way to do these questions where they list all of the distance is to say that displacement = vector sum of all movements, ie. how its been calculated in the last few posts.
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knightrider

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Re: VCE Physics Question Thread!
« Reply #487 on: June 30, 2014, 11:08:45 pm »
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Thanks so much guys and girls it makes so much more sense now love you guys  :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) ::)

knightrider

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Re: VCE Physics Question Thread!
« Reply #488 on: July 09, 2014, 08:16:41 pm »
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Can anyone help with this question


PB

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Re: VCE Physics Question Thread!
« Reply #489 on: July 09, 2014, 10:56:41 pm »
+1
Can anyone help with this question
Well the instantaneous velocity would require you to find the gradient at point t=35s.  So that requires differentiation - which is basically impossible because you have no formula to differentiate!
In this case, I reckon you should try draw a tangent from the point t=35s and find the gradient of that line :P
A bit of a shoddy technique but its VCE Physics. Besides, VCAA is unlikely to give this kind of question because there would be too much variation in the answers obtained. Hope that helps
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Re: VCE Physics Question Thread!
« Reply #490 on: July 09, 2014, 11:58:11 pm »
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Thanks
how would you do this question

Rishi97

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Re: VCE Physics Question Thread!
« Reply #491 on: July 10, 2014, 11:29:35 am »
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A lamp purchased in the USA is designed to operate at optimum efficiency when it is connected to an AC supply that has a peak voltage of 170 V and a frequency of 60 Hz. The lamp has an operating resistance of 100 ohms

What is the peak power produced by this lamp?
I know the peak power = I peak x V peak but this isn't working for me

Thanks :)
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Re: VCE Physics Question Thread!
« Reply #492 on: July 10, 2014, 12:02:30 pm »
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Thanks
how would you do this question
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.
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Re: VCE Physics Question Thread!
« Reply #493 on: July 10, 2014, 12:14:40 pm »
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A lamp purchased in the USA is designed to operate at optimum efficiency when it is connected to an AC supply that has a peak voltage of 170 V and a frequency of 60 Hz. The lamp has an operating resistance of 100 ohms

What is the peak power produced by this lamp?
I know the peak power = I peak x V peak but this isn't working for me

Thanks :)
Just wondering what the answer was?
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Rishi97

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Re: VCE Physics Question Thread!
« Reply #494 on: July 10, 2014, 12:28:25 pm »
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Just wondering what the answer was?

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