• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Physics marathon (hsc) (4 Viewers)

Keelan134

Member
Joined
Nov 24, 2011
Messages
178
Gender
Male
HSC
2012
haha damn, I guess the complexity of the working will depend on the mark allocations but yeah you're right
btw after checking Keelan134's work, final answer should be 6.52 x 10^8 J, which is pretty close to my answer hahaha
What happened during my calculations?
 

Keelan134

Member
Joined
Nov 24, 2011
Messages
178
Gender
Male
HSC
2012
Easy, W = Fd
= (200*9.8)*350000
= 6.86 x 10^8 J

Discuss Einstein's and Planck's differing views of scientific research and its place in society and politics. (4 marks)
Planck was a bonafide nationalist, who believed in his country and would work to benefit it, even if that meant scientifically through his occupation. Signed the manifesto of 93 Intellectuals supporting the role of Germany in the war. Possibly in Planck's fear of persecution that he argued that Scientific research shouldnt be removed from politics and society.

Einstein in contrast was a pacifist. He announced that Scientific should benefit humanity, not contribute to their death, and subsequently became a Swiss citizen, and eventually emigrating to America where he stated his appreciation of it's "meritocracy" in which individuals could "say and do as they please, without social barriers, encouraging the individual to be more creative".

(That last section brought back fond memories of Belonging :( )
 

Keelan134

Member
Joined
Nov 24, 2011
Messages
178
Gender
Male
HSC
2012
can you use that equation??
If it is parabolic flight launching and landing at the same altitude. I showed my Science coordinator when I found the equation and he said it is derived from our suvat equations somehow, so I guess we can. The only reason I used it is because I didn't have enough information to use any other formulas.
 

barbernator

Active Member
Joined
Sep 13, 2010
Messages
1,439
Gender
Male
HSC
2012
Here are some more perceptive questions.

a) A satellite was launched from Earth to Mars. What is the optimal point of interception of the satellite with mars' orbit in relation to where it was originally launched from earth to ensure minimal fuel is required for the journey? (2 marks)

b) A satellite is going to use the slingshot effect to gain maximum speed towards a distant star. At what point within the earth's orbit (see diagram) would it be optimal for the spacecraft to use the slingshot effect, and from what direction should it approach and what direction should it leave to maximise speed (assume the spacecraft can approach from anywhere) (3 marks)

View attachment 26879

c) Calculate the centripetal force on a satellite travelling at a velocity of 40km/h and a radius of 3x10^6 km around a planet of mass 5x10^20 kg.
bump.
 

J-Wang

Member
Joined
May 28, 2012
Messages
120
Gender
Male
HSC
2012
In the current generation prac, HOW were variables held constant? I came across a question which asked this. I knew which variables were held constant, but I didn't know HOW they were held constant. Anyone know?
 

LlamaBoi

Member
Joined
Oct 23, 2012
Messages
31
Gender
Male
HSC
2012
can you use that equation??
You can but you probably shouldnt if you dont understand why you're using it
What I would do to be safe is get an expression for t and then sub it into another equation

i.e. x = Vtcos()
75 = Vtcos(23.2)
t = 75/Vcos(23.2)
then sub that into y = Vsin()t - 0.5at^2
0 = Vsin(23.2)*(75/Vcos(23.2) - (9.8/2)(75^2/V^2cos^2(23.2))
then two of the V's cancel out and after some simple algebra you get V^2 = 1014.952
so V = 31.86
 

barbernator

Active Member
Joined
Sep 13, 2010
Messages
1,439
Gender
Male
HSC
2012
In the current generation prac, HOW were variables held constant? I came across a question which asked this. I knew which variables were held constant, but I didn't know HOW they were held constant. Anyone know?
If the constant variable is distance magnet was moved inwards, just say "we marked out a distance to move the magnet" or something like that.
 

leesh95

Member
Joined
Nov 7, 2011
Messages
487
Gender
Undisclosed
HSC
2013
A satellite of mass 750 kg is changed from an orbital radius of 6000km to a geostationary orbit of radius 5*10^4
Calculate the change in its gravitational potential energy?
 

Danstar2

Member
Joined
Sep 19, 2012
Messages
65
Gender
Male
HSC
2012
A satellite of mass 750 kg is changed from an orbital radius of 6000km to a geostationary orbit of radius 5*10^4
Calculate the change in its gravitational potential energy?
-GMm/r

P.E at= 6*10^6m
minus
P.E at= 5*10^4m

Thus: -(6.67*10^-11 x 6*10^24 x 750)/12.38*10^6 - [-(6.67*10^-11 x 6*10^24 x 750)/5.638*10^4]

= -2.424*10^10 - ( -5.324*10^12)

= 5.3*10^12j

Can someone tell me if this is wrong or not? Not sure myself.
 
Last edited:

shongaponga

Member
Joined
Feb 15, 2012
Messages
125
Gender
Male
HSC
2012
-GMm/r

P.E at= 6*10^6m
minus
P.E at= 5*10^4m

Thus: -(6.67*10^-11 x 6*10^24 x 750)/12.38*10^6 - [-(6.67*10^-11 x 6*10^24 x 750)/5.638*10^4]

= -2.424*10^10 - ( -5.324*10^12)

= 5.3*10^12j

Can someone tell me if this is wrong or not? Not sure myself.
I think your working is fine, but note that the geostationary orbit of 5 x 10^4 > orbital radius of 6000. So the only thing i'm wondering is that maybe your answer should be a negative value? Perhaps it might not make a difference.
 

Danstar2

Member
Joined
Sep 19, 2012
Messages
65
Gender
Male
HSC
2012
Yeah, well now I'm not sure whether I needed to include the radius of the earth in the answer either. I just looked at a worked example of a geostat orbit at 35800km and the solution did not take into account the radius. However, on the contrary I am 99% sure that the distance was from the centre of the mass. So now I'm not sure which to believe, the textbook answer or the theory...
 

barbernator

Active Member
Joined
Sep 13, 2010
Messages
1,439
Gender
Male
HSC
2012
Yeah, well now I'm not sure whether I needed to include the radius of the earth in the answer either. I just looked at a worked example of a geostat orbit at 35800km and the solution did not take into account the radius. However, on the contrary I am 99% sure that the distance was from the centre of the mass. So now I'm not sure which to believe, the textbook answer or the theory...
the distance/radius is measured from the centre of mass. In saying "a geostationary orbit is 35800km" this value already takes into account the radius of the earth, as the satellite is 35800km above the dead centre of the earth. In calculation questions they will be more specific as to whether the distance stated is above the earths surface or above the centre of the earth.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 4)

Top