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HSC Physics Marathon 2013-2015 Archive (1 Viewer)

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Kaido

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re: HSC Physics Marathon Archive

OK:

given g=9.8, G=6.67x10^-11 and Mass of earth=6x10^24 and let m2=1kg (in this case) <-sub in w/e value in the exam

g=Gm1m2/r^2
r^2=(6.67x10^-11x6x10^24)/9.8
r=6387.1km
 
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Fizzy_Cyst

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re: HSC Physics Marathon Archive

Correct working, but I end up with a value of 6390km?

New Question:

Determine the work required to accelerate an electron from rest to 0.98c (3 marks)
 

Kaido

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re: HSC Physics Marathon Archive

W=Energy (kinetic i think)

E=m(0)xc^2x(1-v^2/c^2)^-0.5
=9.109×10^-31x(3x10^8)^2x((1-0.98^2)^0.5-1)
=4.1197x10^-13J

Edit: Forgot two negative signs...
 
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Fizzy_Cyst

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re: HSC Physics Marathon Archive

W=Energy (kinetic i think)

E=m(0)xc^2x(1-v^2/c^2)^-0.5
=9.109×10^-31x(3x10^8)^2x((1-0.98^2)^0.5-1)
=4.1197x10^-13J

Edit: Forgot two negative signs...


I think you forgot to halve it ;)

But correct process :)

Here is one my tutor students struggled with in their examination

New Question: (There was a diagram to go with it, but it makes sense without it)


Two clocks are started simultaneously.
A rocketship instantaneously accelerates to 0.6c as the clocks are started.
At the instant in time that the observer on the ground measures the rocketship to have travelled 5.4x10^10m, how much time has the observer in the rocketship seen elapse on the clock on the ground?
 

Kaido

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re: HSC Physics Marathon Archive

I'm thinking 240s? (will show working after you say it's correcto :) )
 

Fizzy_Cyst

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re: HSC Physics Marathon Archive

That's not the answer, but it is part way there :)
 

Kaido

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re: HSC Physics Marathon Archive

Ahh dayum, didn't read that last part (sneaky...)

Then i reckon it's 300s

(second guessing myself)

Edit: It's 192, shown by integrand below
 
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colinrocks95

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re: HSC Physics Marathon Archive

Question:

3 moons, X, Y and Z are in orbit around the same planet. The moons have identical orbital speeds, but masses of M, 9M and 25M respectively. Determine the ratio of their orbital radii and justify your answer.
Since each moon orbits the same planet, M_planet is the same for each of them. Orbital velocity is independent of the mass of the orbiting object. Since each moon has the same orbital velocity, they all have THE SAME ORBITAL RADII, i.e. the ratio is 1:1:1. This is because Kepler's formula (on the data sheet) is r^3/T^2 = G*M_planet / 4*pi^2. Note: equal orbital velocities mean equal orbital periods (T), which is the time it takes for one complete orbit.

My question is:
Planet A has half the radius of planet B. Both planets have different masses. What is the ratio of their masses if the speed of an object orbiting planet A, at an orbital radius equal to the radius of planet B, is equal to the minimum speed required to launch the object from the surface of planet B?


By the way, I got good at maths and physics from the HSC programs at Sci School. The presenter is funny and explains things really well. You should check them out.
 

InteGrand

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re: HSC Physics Marathon Archive

I think you forgot to halve it ;)

But correct process :)

Here is one my tutor students struggled with in their examination

New Question: (There was a diagram to go with it, but it makes sense without it)


Two clocks are started simultaneously.
A rocketship instantaneously accelerates to 0.6c as the clocks are started.
At the instant in time that the observer on the ground measures the rocketship to have travelled 5.4x10^10m, how much time has the observer in the rocketship seen elapse on the clock on the ground?

Let and .

When the observer on the ground measures the spaceship to have travelled the distance , the time elapsed on his/her ground clock is

.

Therefore, the time that the ground observer sees elapsed on the spaceship clock at this instant is

.

When the spaceship observer sees his/her spaceship clock have this time of elapsed, he/she sees the ground clock have the following time elapsed:

.

Now, plugging in the values given in the question, the answer is





.
 

Kaido

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re: HSC Physics Marathon Archive

Yeah it was 192 (was tossing up between either 192 or 300, forgot that relativistic time is slower)
 

astroman

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re: HSC Physics Marathon Archive

How do you calculate the distance in a light year?
 

Kaido

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re: HSC Physics Marathon Archive

@astro cx60x60x24x365
 

astroman

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how do u calculate the distance to proxima centauri using parallax?
 

astroman

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dw i got it,

d=1/0.772
d=1.295pc
1 pc=3.2616 ly
1.295pc=3.2616 x 1.295
=~4.2ly
 

Kaido

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ohh i see, astrophysics seems interesting, but Q2Q still the boss :D
 
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