HSC Physics Marathon 2013-2015 Archive (2 Viewers)

Anvaeon · Feb 20, 2013

Re: 2013 HSC physics marathon

iBibah said:
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]

someth1ng · Feb 20, 2013

Re: 2013 HSC physics marathon

Anvaeon said:
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.

iBibah · Feb 23, 2013

Re: 2013 HSC physics marathon

someth1ng said:
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.

Fizzy_Cyst · Feb 24, 2013

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.

someth1ng · Feb 24, 2013

re: HSC Physics Marathon Archive

Fizzy_Cyst said:
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?

Fizzy_Cyst · Feb 24, 2013

re: HSC Physics Marathon Archive

someth1ng said:
That's just calculating the change in GPE, isn't it?

Nope

Sy123 · Feb 28, 2013

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$

Fizzy_Cyst · Mar 24, 2013

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.

Sy123 · Mar 24, 2013

re: HSC Physics Marathon Archive

Fizzy_Cyst said:
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.

study1234 · Mar 24, 2013

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.

Fizzy_Cyst · Mar 25, 2013

re: HSC Physics Marathon Archive

Sy123 said:
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

bleakarcher · Mar 25, 2013

re: HSC Physics Marathon Archive

Fizzy_Cyst said:
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?

Fizzy_Cyst · Mar 25, 2013

re: HSC Physics Marathon Archive

bleakarcher said:
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

someth1ng · Mar 25, 2013

re: HSC Physics Marathon Archive

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

Sy123 · Apr 8, 2013

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$

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

yasminee96 · Apr 10, 2013

re: HSC Physics Marathon Archive

Sy123 said:
$$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$

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

q/m = 1 x10^5
mv^2 /r = qvB (theta = 90degrees)
q/m = v/rB

v/r = q/m x B
v/r =1 x 10^5 x 10
r/v = 1 x 10^-6
distance/speed = time
therefore time = 1x10^-6 seconds

maybe?

yasminee96 · Apr 10, 2013

Re: 2013 HSC physics marathon

Immortalp00n said:
The slingshot effect, defined as the increase in velocity given to a spacecraft because it enters the gravitational field of a planet as it passes by, is a technque used by Space agencies in accordance with physicists in order to take full advantage of a planets rotational motion in order to increase a particle's ( spacecraft) final speed.
The slingshot effect is used by many spacecraft which travel within and beyond the solar system.
A spacecraft passes close to a a planet such that its gravity pulls the spacecraft into an arc or circular motion.It leaves with the same speed relative to the planet, but its speed is increased when viewed from an alternate frame of reference such as the Sun.
The acquired speed is large enough so that the spacecraft can travel away from the planet.
Comparatively, a ball thrown from a person who is stationary would have speeds less than a ball thrown from a person who is running at an increasing or stationary speed. Thus the slingshot effect, also known as gravity assist trajectory is used in order to attain a significant change in speed and direction despite the very little expenditure of fuel, improving flight efficiency.

Should possibly consider writing this as "increase in velocity, relative to the Sun", because relative to the planet it passes, no change in velocity is experienced.

Sy123 · Apr 10, 2013

re: HSC Physics Marathon Archive

yasminee96 said:
q/m = 1 x10^5
mv^2 /r = qvB (theta = 90degrees)
q/m = v/rB

v/r = q/m x B
v/r =1 x 10^5 x 10
r/v = 1 x 10^-6
distance/speed = time
therefore time = 1x10^-6 seconds

maybe?

That looks correct to me, nice work.

psychotropic · Apr 15, 2013

re: HSC Physics Marathon Archive

Almost a correct solution

However, for one complete revolution the distance is r2pi.
It's a minor error, probs only lose 2 marks
Although in the future try not to just use formulae without thinking

anomalousdecay · Apr 19, 2013

re: HSC Physics Marathon Archive

yasminee96 said:
q/m = 1 x10^5
mv^2 /r = qvB (theta = 90degrees)
q/m = v/rB

v/r = q/m x B
v/r =1 x 10^5 x 10
r/v = 1 x 10^-6
distance/speed = time
therefore time = 1x10^-6 seconds

maybe?

isn't it :

t=2(r)(pi)/v

t= 2(10^-4)(pi) = 6.2831*10^-4 seconds

??

HSC Physics Marathon 2013-2015 Archive (2 Viewers)

New Member

Retired Nov '14

Well-Known Member

Owner @ Sigma Science + Phys Goat

Retired Nov '14

Owner @ Sigma Science + Phys Goat

This too shall pass

Owner @ Sigma Science + Phys Goat

This too shall pass

Member

Owner @ Sigma Science + Phys Goat

Active Member

Owner @ Sigma Science + Phys Goat

Retired Nov '14

This too shall pass

Active Member

Active Member

This too shall pass

Member

Premium Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)