# Thread: HSC Physics Marathon 2013-2015 Archive

1. ## Re: 2013 HSC physics marathon

Originally Posted by bleakarcher
Yeah, I just meant applying the mass dilation formula to F=ma:
a=F/(m/sqrt(1-v^2/c^2))=F*sqrt(1-v^2/c^2)/m
As v->c, a->0.
thank you for that.

2. ## Re: 2013 HSC physics marathon

If our galaxy, the Milky Way, is 20 kiloparsecs or 65 000 light-years in radius, calculate how fast a spacecraft would need to travel so that its occupants could travel right across it in 45 years.
Interesting question.

3. ## Re: 2013 HSC physics marathon

Fuuuuuar! Just noticed some of my tutor students pwning this thread. Well done you three

4. ## Re: 2013 HSC physics marathon

QUESTION: Explain why all low earth orbit satellites will eventually fall to the Earth’s surface.

5. ## Re: 2013 HSC physics marathon

Originally Posted by iBibah
QUESTION: Explain why all low earth orbit satellites will eventually fall to the Earth’s surface.
All Low Earth Orbit (LEO) satellites will eventually fall to the Earth's surface due to atmospheric drag. Because LEOs are typically 160-2000km above the surface of the Earth, the satellite tends to collide with air molecules, this creates friction and thus kinetic energy is converted into heat and sound energy, thereby further losing kinetic energy and in time, altitude.

6. ## Re: 2013 HSC physics marathon

$A projectile is fired such that the maximum height is 100m, and the range is 100m$

$Find the angle and the velocity by which the projectile must be fired$

7. ## Re: 2013 HSC physics marathon

Originally Posted by Sy123
$A projectile is fired such that the maximum height is 100m, and the range is 100m$

$Find the angle and the velocity by which the projectile must be fired$
Just a note, am I allowed to ask questions which need students to solve simultaneous equations? Surely general maths students can do that.

=======================

$It is given that \ \ \ 2 \times \sin \theta \times \cos \theta = \sin(2\theta)$

$Prove that if a projectile is fired under no air resistance. Then the angle that will maximise the range of the projectile is \ \ 45^{\circ}$

8. ## Re: 2013 HSC physics marathon

Originally Posted by Sy123
Just a note, am I allowed to ask questions which need students to solve simultaneous equations? Surely general maths students can do that.

=======================

$It is given that \ \ \ \sin \theta \times \cos \theta = \sin(2\theta)$

$Prove that if a projectile is fired under no air resistance. Then the angle that will maximise the range of the projectile is \ \ 45^{\circ}$
You may want to fix up the given equation, you've done a typo I think.

9. ## Re: 2013 HSC physics marathon

Originally Posted by RealiseNothing
You may want to fix up the given equation, you've done a typo I think.
haha oops thanks for the heads up

10. ## Re: 2013 HSC physics marathon

Q: Calculate the effective value of 'g' at the equator. (Given g=9.81ms^-2)

11. ## Re: 2013 HSC physics marathon

Originally Posted by iBibah
Q: Calculate the effective value of 'g' at the equator. (Given g=9.81ms^-2)
A. The equation to find the value of 'g' is ge= GM/r^2. The radius of the Earth is only slightly bulged at the equator (and flattened at the poles), resulting in only a slight increase in radius and variation from the given "g=9.81ms^-2". The resulting acceleration due to gravity at the equator is about 9.782ms^-2 (sea level).

Q. Describe how the motor effect is used in a loudspeaker. [3 marks]

12. ## Re: 2013 HSC physics marathon

Originally Posted by Anvaeon
A. The equation to find the value of 'g' is ge= GM/r^2. The radius of the Earth is only slightly bulged at the equator (and flattened at the poles), resulting in only a slight increase in radius and variation from the given "g=9.81ms^-2". The resulting acceleration due to gravity at the equator is about 9.782ms^-2 (sea level).

Q. Describe how the motor effect is used in a loudspeaker. [3 marks]
It clearly said "calculate" - an response like that would get 0.

13. ## Re: 2013 HSC physics marathon

Originally Posted by someth1ng
It clearly said "calculate" - an response like that would get 0.
This.

Radius of earth = 6378km (probably should have given this)

Therefore velocity at equator: v = (2*pi*6378000)/24*60*60

Centripetal acceleration = v^2/r = 462.82^2/6378000=0.03373 ms^-2

Therefore effective value of 'g' at the equator is reduced by 0.033 ms^-2, and the resultant effective value of 'g' is 9.777 ms^-2.

14. ## re: HSC Physics Marathon Archive

Nice solution Habibi!

New Question:

A satellite is launched from the equator to a stable orbit of altitude 600km. Determine the work done on the satellite.

15. ## re: HSC Physics Marathon Archive

Originally Posted by Fizzy_Cyst
Nice solution Habibi!

New Question:

A satellite is launched from the equator to a stable orbit of altitude 600km. Determine the work done on the satellite.
That's just calculating the change in GPE, isn't it?

16. ## re: HSC Physics Marathon Archive

Originally Posted by someth1ng
That's just calculating the change in GPE, isn't it?
Nope

17. ## re: HSC Physics Marathon Archive

$A 5T magnetic field directed out of the page changes to 0T in 0.1 seconds.$

$Determine the emf generated in a wire loop of \ \ 2m^2 \ \ in area. Find the magnitude and direction of the current if the resistance of the wire loop is 20 Ohms$

18. ## re: HSC Physics Marathon Archive

If Power loss = v^2/r, then why do we transmit electricity at high voltage. Should this not increase power loss? Explain.

19. ## re: HSC Physics Marathon Archive

Originally Posted by Fizzy_Cyst
Nope
I'm pretty sure it is, isn't it?
If we define work properly, let point A be on the equator and point B be the point 600km away:

$W_{AB} = \int_{A}^{B} F \ dx$

Where F is force, and x is the distance.

By Newton's Law of Universal Gravitation:

$F= \frac{GMm}{x^2}$

$W_{AB} = \int_{A}^{B} \frac{GMm}{x^2} \ dx$

$W_{AB}= \left(\frac{-GMm}{x}\right)_A^B$

$W_{AB}= E_{p_{B}} - E_{p_{A}}$

Where E_p is the gravitational potential energy.

20. ## re: HSC Physics Marathon Archive

You should write it as P=I^2 R. Thus since power is proportional to the square of current, there is less power loss when there is low current. Since V=P/I, for a low current, there is a high voltage. Thus, a step-up transformer is used to increase the voltage for minimal power/energy loss in transmission lines.

21. ## re: HSC Physics Marathon Archive

Originally Posted by Sy123
I'm pretty sure it is, isn't it?
If we define work properly, let point A be on the equator and point B be the point 600km away:

$W_{AB} = \int_{A}^{B} F \ dx$

Where F is force, and x is the distance.

By Newton's Law of Universal Gravitation:

$F= \frac{GMm}{x^2}$

$W_{AB} = \int_{A}^{B} \frac{GMm}{x^2} \ dx$

$W_{AB}= \left(\frac{-GMm}{x}\right)_A^B$

$W_{AB}= E_{p_{B}} - E_{p_{A}}$

Where E_p is the gravitational potential energy.
You are making an assumption

22. ## re: HSC Physics Marathon Archive

Originally Posted by Fizzy_Cyst
Nice solution Habibi!

New Question:

A satellite is launched from the equator to a stable orbit of altitude 600km. Determine the work done on the satellite.
Is it the change in the mechanical energy of the satellite?

23. ## re: HSC Physics Marathon Archive

Originally Posted by bleakarcher
Is it the change in the mechanical energy of the satellite?
Yes

Not only does work need to be done to lift it from the Earth to 600km altitude, work must also be done to make its speed increase to the orbital velocity of an orbit of alitutude 600km.

Think about it, change in Ep is just saying 'lifting it from the surface of Earth and placing at a point 600km away', whereas this satellite is ORBITING, not just sitting there, hence it needs more energy and the amount of extra energy it needs is equal to the change in Ek of the satellite

24. ## re: HSC Physics Marathon Archive

Ahh, forgot about that - that's why I'm not gonna be majoring in Physics :P

25. ## re: HSC Physics Marathon Archive

$A charge has charge-mass ratio \ \ 1 \times 10^{5}$

$The charge travels in a magnetic field of strength \ \ 10 T \ \ perpendicular to its trajectory, the charge moves in a circle$

$Calculate the time the charge takes to complete a full revolution of the circle$

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