September 21, 2017, 05:54:35 am

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#### Jakeybaby

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##### Re: VCE Physics Question Thread!
« Reply #1755 on: December 09, 2016, 09:20:05 pm »
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Significant figures are not assessed VCE physics, which is a shame I guess.

To OP: like Buddster asserted, only round your answer to a sensible amount of numbers.
Yeah for physics it isnt like chemistry, they honestly dont care about sig figs. The sig figs are mostly incorrect in the examiners reports (recently they've been better but pre 2012 there's like 0 care for sig figs)
as long as u dont round too much, (i.e answer is 2.6 and you write 3)
Fair enough, wasn't expecting that haha
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#### Gogo14

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##### Re: VCE Physics Question Thread!
« Reply #1756 on: January 03, 2017, 03:58:25 pm »
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1. When an object is spinning, why does the object continue to move along the circumference if it is accelerating towards the centre. I.e why does centripedal acceleration not cause it to reach the centre?Because its accelerating toward the centre, shouldnt it eventuall reach the centre?
2. In the photo, why is the reaction force acting on the cyclist at an angle. I thought the reaction force was always perpendicular?

#### Syndicate

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##### Re: VCE Physics Question Thread!
« Reply #1757 on: January 03, 2017, 04:11:23 pm »
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1. When an object is spinning, why does the object continue to move along the circumference if it is accelerating towards the centre. I.e why does centripedal acceleration not cause it to reach the centre?Because its accelerating toward the centre, shouldnt it eventuall reach the centre?
2. In the photo, why is the reaction force acting on the cyclist at an angle. I thought the reaction force was always perpendicular?

1. Centrifugal force (it is an apparent force) works opposite to the centripetal force, which draws the rotating body out the centre.

2. I think it is a mistake because normal force is always perpendicular to the ground (the book would only be right, if they are trying to show the bike on a banked track).
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#### homosapien

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##### Re: VCE Physics Question Thread!
« Reply #1758 on: January 05, 2017, 12:07:14 am »
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Could someone please explain the trend in size of the weight force (Fg) and normal force (FN) as a ball bounces / goes up and down?
Thanks

#### Syndicate

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##### Re: VCE Physics Question Thread!
« Reply #1759 on: January 05, 2017, 10:20:01 am »
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Could someone please explain the trend in size of the weight force (Fg) and normal force (FN) as a ball bounces / goes up and down?
Thanks

As the:
- ball is bouncing down, only gravity exists (since it is free falling).
- ball touches the ground: Fg = Fn
- ball compresses: Fn > Fg
- ball exits it's compression stage, however, is still touching the ground: Fg = Fn
- ball going back up: Fn = 0, and only Fg exists
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#### Xandorious

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##### Re: VCE Physics Question Thread!
« Reply #1760 on: January 11, 2017, 11:53:25 am »
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A spacecraft leaves Earth to travel to the Moon. How far from the centre of the Earth is the spacecraft when it experiences a net force of zero?

#### wyzard

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##### Re: VCE Physics Question Thread!
« Reply #1761 on: January 11, 2017, 12:20:10 pm »
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A spacecraft leaves Earth to travel to the Moon. How far from the centre of the Earth is the spacecraft when it experiences a net force of zero?

That is a pretty intense question I'll give you some hints and pointers.

The gravitational force by the Earth and Moon is at opposite directions, and the closer you get to one planet the stronger the gravitational pull. So there must be a point somewhere in between that will have the forces cancelling each other out.

To find the distance, let x be the distance from the center of Earth. Use this x, set up the equation of forces from the Earth and Moon using Newton's law of gravity, then solve for x.
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#### Rathin

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##### Re: VCE Physics Question Thread!
« Reply #1762 on: January 11, 2017, 01:27:52 pm »
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A spacecraft leaves Earth to travel to the Moon. How far from the centre of the Earth is the spacecraft when it experiences a net force of zero?

Fnet (from moon) = GM(moon)m(satellite)/r(moon)^2
Fnet (from earth = GM(earth)m(satellite)/r(earth)^2

Where r(moon) is the distance from moon centre to the satellite
and r(earth) is the distance from the earth centre to the satellite

Hence at the point where net force  on satellite  = 0

We will equate the two formulas;

m(moon)/r(moon)^2 = m(earth)/r(earth)^2

and since  r(moon) = 3.84 x 10^8 - r(earth)

m(moon)/ (3.84 x 10^8 - r(earth))^2 =  m(earth)/r(earth)^2

sub in relevant numbers

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#### Xandorious

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##### Re: VCE Physics Question Thread!
« Reply #1763 on: January 12, 2017, 03:16:46 pm »
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@Rathin

Hi, i'm glad that you got the answer that I got but the answer in my book states 3.02 * 10^8 m

#### Gogo14

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##### Re: VCE Physics Question Thread!
« Reply #1764 on: January 19, 2017, 01:00:13 pm »
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How do you do quesiton 9, graph relates to question
http://imgur.com/6lYLwa2

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##### Re: VCE Physics Question Thread!
« Reply #1765 on: January 19, 2017, 11:24:00 pm »
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@Rathin

Hi, i'm glad that you got the answer that I got but the answer in my book states 3.02 * 10^8 m
Unless we both made the same mistake, I got that answer too so it may be a book error

How do you do quesiton 9, graph relates to question
http://imgur.com/6lYLwa2

Hi, I'm a bit rusty on this, so someone correct me if I'm wrong.
I'm assuming at the start of the question it means 6000km, not 600km as that's what's shown on the graph.
J = Nm
Change in the energy is the change in kinetic energy, final kinetic energy is zero, so the initial kinetic energy is the change in kinetic energy.

Change in kinetic energy is the area under the graph, so just the number of squares * value of squares (in this case 2*(1/2)*10^6 = 10^6)
16 squares * 10^6
1.6*10^7 Joules

Or you could multiply each of the values and calculate the difference aka change in energy
Initial energy = 11*6.0*10^6 = 6.6*10^7
Final energy = 6.3*8.0*10^6 = 5.0*10^7
Difference = 1.6*10^7 Joules

Hope this helps
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#### Gogo14

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##### Re: VCE Physics Question Thread!
« Reply #1766 on: January 23, 2017, 03:19:18 pm »
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Unless we both made the same mistake, I got that answer too so it may be a book error

Hi, I'm a bit rusty on this, so someone correct me if I'm wrong.
I'm assuming at the start of the question it means 6000km, not 600km as that's what's shown on the graph.
J = Nm
Change in the energy is the change in kinetic energy, final kinetic energy is zero, so the initial kinetic energy is the change in kinetic energy.

Change in kinetic energy is the area under the graph, so just the number of squares * value of squares (in this case 2*(1/2)*10^6 = 10^6)
16 squares * 10^6
1.6*10^7 Joules

Or you could multiply each of the values and calculate the difference aka change in energy
Initial energy = 11*6.0*10^6 = 6.6*10^7
Final energy = 6.3*8.0*10^6 = 5.0*10^7
Difference = 1.6*10^7 Joules

Hope this helps

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##### Re: VCE Physics Question Thread!
« Reply #1767 on: January 23, 2017, 04:13:45 pm »
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Ah I see, haven't done many questions on this and don't think I've seen any on an exam (so very rusty, sorry)
First of all I misread the question partially as it's in orbit 600km from earth's surface, and the radius of earth is approx 6.4*10^3km, the satellite is approx 7*10^3km or 7*10^6m from the centre of the earth.
So the energy per kg of the satellite is 7 squares *10^6 (value of each of the squares) = 7*10^6J
Total change in energy for the satellite is 7*10^6 J * 240kg = 1.7*10^9J
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#### Gogo14

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##### Re: VCE Physics Question Thread!
« Reply #1768 on: January 23, 2017, 04:44:14 pm »
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Ah I see, haven't done many questions on this and don't think I've seen any on an exam (so very rusty, sorry)
First of all I misread the question partially as it's in orbit 600km from earth's surface, and the radius of earth is approx 6.4*10^3km, the satellite is approx 7*10^3km or 7*10^6m from the centre of the earth.
So the energy per kg of the satellite is 7 squares *10^6 (value of each of the squares) = 7*10^6J
Total change in energy for the satellite is 7*10^6 J * 240kg = 1.7*10^9J

But why do you multiply 7*10^6 by 240 kg? Isnt the graph a force vs distance graph not a gravitational field strength vs  distance graph

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##### Re: VCE Physics Question Thread!
« Reply #1769 on: January 23, 2017, 05:00:27 pm »
+1
But why do you multiply 7*10^6 by 240 kg? Isnt the graph a force vs distance graph not a gravitational field strength vs  distance graph

It says on the graph that it's the force acting on 1kg, aka the gravitational field strength, which is why you have to multiply the weight.
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