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Author Topic: VCE Physics Question Thread!  (Read 603408 times)  Share 

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lzxnl

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Re: VCE Physics Question Thread!
« Reply #855 on: February 15, 2015, 08:57:44 pm »
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Hey guys. I just started gravity and i'm already stuck on this question 3a. What does in terms of the radius on the moon mean. Am i supposed to know the radius of the moon already from my textbook?
The force is 4x weaker at the new radius so the radius is 4x greater? Does this mean its a variation question. Am i supposed to introduce a constant to solve it?

Spoiler

The gravitational force depends on 1/r^2, so if the force has quartered, the radius has doubled
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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #856 on: February 15, 2015, 09:48:20 pm »
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So because I know how much the force has decreased by, I can then just solve it like so: 







? And also how do you do 3b then?
« Last Edit: February 15, 2015, 09:55:55 pm by Floatzel98 »
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #857 on: February 15, 2015, 10:20:31 pm »
+2
More like

F = k/r^2
F1/F2 = (r2/r1)^2
Force ratio is 4 as the second force F2 is a quarter of the original one, so 4 = (r2/r1)^2
r2/r1 = 2, r2  = 2 r1

As for the next one, three Moon radii above the surface = four Moon radii from the centre of the Moon => quadruple the radius => force shrinks to 1/16 of original
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orgekas

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Re: VCE Physics Question Thread!
« Reply #858 on: February 17, 2015, 04:37:43 pm »
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Hi guys,

I've got the following investigation(see photo) coming up and I have no idea how to do it.
Any ideas?
The aim is to design an experiment that will prove the formula F= mv^2/r for a body moving in a uniform circle.
I hope that you can read the attachment for more information. I f you can't, TELL ME to upload a better version.
The teacher said that we should use graphs.

Thank you.

odeaa

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Re: VCE Physics Question Thread!
« Reply #859 on: February 17, 2015, 09:10:19 pm »
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Need help with this question, the 'worked' solutions are useless.
I also don't understand the whole dilation of gravity thing
Thanks

VCE Class of 2015

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Stevensmay

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Re: VCE Physics Question Thread!
« Reply #860 on: February 18, 2015, 12:13:08 am »
+3
Need help with this question, the 'worked' solutions are useless.
I also don't understand the whole dilation of gravity thing
Thanks

(Image removed from quote.)

Torque is defined as a force multiplied by the perpendicular distant from that point to the force. (More technically it's the cross product of radius and force, this is why it's perpendicular). In the real world it's effectively a force that makes something want to rotate. In the picture the 20kg force wants to make the beam rotate around point P.

In our case the force from the 20kg mass is 20*9.8N (mg).



So we've decided that a torque is just a measure of how much something wants to rotate. Higher torque means something is wanting to rotate more. In the image above, would the left hand side 20kg mass produce a torque smaller than, equal to or greater than the torque on the right hand side?

Spoiler
Equal to, because only the perpendicular distance matters and it is the same in each diagram. Similarly, if we put the 20kg force above point P it would have a torque of 0, as it is not trying to rotate the beam.

So we now what our perpendicular distance is, it's just a matter of going T = r*F to get an answer of (hopefully, if I read the question right) 588Nm.

If this doesn't make any sense my apologies.

With the second part, I can't remember if you do summation of torques and forces.

Floatzel98

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Re: VCE Physics Question Thread!
« Reply #861 on: February 20, 2015, 04:00:50 pm »
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I found 7a, but don't know what to do for 7b or 8.

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Kel9901

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Re: VCE Physics Question Thread!
« Reply #862 on: February 20, 2015, 06:15:07 pm »
+1
I found 7a, but don't know what to do for 7b or 8.

Spoiler

Let the distance that must be calculated be r
m is the mass of the asteroid

F(asteroid, Alpha)=GMm/r^2
F(asteroid, Beta)=G(0.01M)m)/(R-r)^2

F(asteroid, Alpha)/F(asteroid, Beta)=100(R-r)^2/r^2=8100
(R-r)^2/r^2=81
(R-r)/r=9 (remember that r and R are both positive, and that R>r)
R-r=9r
10r=R
r=0.1R
s=change in displacement for physics
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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #863 on: February 20, 2015, 07:21:57 pm »
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Thank you. Makes a lot more sense now
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Kel9901

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Re: VCE Physics Question Thread!
« Reply #864 on: February 20, 2015, 08:54:13 pm »
+1
For question 8:
GMm/r^2=G(0.01M)m/(R-r)^2
1/r^2=0.01/(R-r)^2
100/r^2=1/(R-r)^2
r^2/100=(R-r)^2
r/10=R-r
1.1r=R
r=0.91R
s=change in displacement for physics
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lolaishappy

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Re: VCE Physics Question Thread!
« Reply #865 on: February 21, 2015, 02:53:06 pm »
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Can anyone confirm answers to questions on the 2014 vcaa exam for Q2 about the spring?
Newb coming through

Kel9901

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Re: VCE Physics Question Thread!
« Reply #866 on: February 22, 2015, 08:25:18 am »
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Can anyone confirm answers to questions on the 2014 vcaa exam for Q2 about the spring?
Yep, I did it last year and got it right (acc. to statement of marks)

a) k=F/x=mg/x pretty simple show that

b) 80 cm

c) Kinetic energy not included

d) 2 ms^-1
s=change in displacement for physics
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lolaishappy

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Re: VCE Physics Question Thread!
« Reply #867 on: February 22, 2015, 08:32:50 am »
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Yep, I did it last year and got it right (acc. to statement of marks)

a) k=F/x=mg/x pretty simple show that

b) 80 cm

c) Kinetic energy not included

d) 2 ms^-1
Thank you :)
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paper-back

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Re: VCE Physics Question Thread!
« Reply #868 on: February 25, 2015, 08:11:34 pm »
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For questions such as the following:
A person pushes a 14.5KG mower at a constant speed with a force of 80N directed along the handle, which is at an angle of 45 degrees to the horizontal
(A) Find the normal force

Do we need to go; .
9.8x14.5+sin(45)x80?

Cosec

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Re: VCE Physics Question Thread!
« Reply #869 on: February 25, 2015, 08:40:06 pm »
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For questions such as the following:
A person pushes a 14.5KG mower at a constant speed with a force of 80N directed along the handle, which is at an angle of 45 degrees to the horizontal
(A) Find the normal force

Do we need to go; .
9.8x14.5+sin(45)x80?

Normal force always acts perpendicular to the direction of motion. In this case, if its traveling along flat ground, the net force vertically is 0 as the mower isnt moving up or down, but rather left and right. Thus, using f=mg, f=14.5x10 (take gravity as 10m/s^2) gives us 145N (force of gravity acting downwards). Thus the reaction (normal) force is 145N in the opposite direction (acting upwards). In this case, the ground is pushing the mower with a force of 145N. Check the answers because i might of interpreted the question wrong (next time post a photo of the actual whole question with diagram if there is, makes it a bit easier).
« Last Edit: February 25, 2015, 08:49:55 pm by Cosec »