HSC Physics Marathon 2013-2015 Archive (3 Viewers)

[Zeref]

Re: HSC Physics Marathon 2014

Yeah, can save you heaps of time Fw=Fg=Fc=Fnet in orbit!

I'm glad you haven't done M&G, it will give someone else a chance to answer a question, you tank! Hahaha

lolwat.

Tank? I wish lol. Post a troll projectile motion question and I'll be sitting there for 20 minutes getting nowhere. The first question you posted was so much more challenging than the others, I'll probably attempt the question when I have time

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

Assuming length of second wire = length of first wire,
F = 0.05N (force exerted on first wire perpendicularly towards balance)

Now, old weight w1 = mg = (8.154)x 9.8 --- assuming g = 9.8
Therefore new weight w2 = w1 + 0.05N

Thus, new reading = w2/9.8 ~~ 8.159 g

Tell me how I did *my answer relies on the fact that scales take weight as an input then divide by 9.8 to give mass an output.

Correct method, you have just made one very common mistake in the working

lolwat.

Tank? I wish lol. Post a troll projectile motion question and I'll be sitting there for 20 minutes getting nowhere. The first question you posted was so much more challenging than the others, I'll probably attempt the question when I have time

I will think of a hard question tomorrow

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

Ok, couldn't wait til tomorrow. Not heaps hard, but tricky.

New Question:

Whilst playing tennis, John serves a ball with a velocity of 45ms-1 at an angle of 6.5degrees below the horizontal. The height of the ball as it leaves his racquet is 2.45m above the ground. The net is 12.2m away and is of height 0.92m.
Determine whether the ball clears the net.

 

[QZP]

Re: HSC Physics Marathon 2014

Correct method, you have just made one very common mistake in the working



I will think of a hard question tomorrow

I have scrambled my brain for long enough. What's the error? :S

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

Convert g --> kg

 

[QZP]

Re: HSC Physics Marathon 2014

troll LOL Would I lose a mark in the hsc for that?

 

[Zeref]

Re: HSC Physics Marathon 2014

troll LOL Would I lose a mark in the hsc for that?

I think so, you need the correct SI units. Not converting and substituting into formula's will give you a wrong answer. (an exception would be the special relativity questions, as long as you keep your units constant ofc)

 

[QZP]

Re: HSC Physics Marathon 2014

Ah yes, you are right. I thought I only made the mistake in my final answer, but then I realised different. Thanks

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

troll LOL Would I lose a mark in the hsc for that?

Yep, 100%. I dont mean the final answer (as that reading would be in grams), but rather the original reading of 8.154g needs to be converted to kg when you determine Weight, should come out to be 0.0799N, then continue your approach.

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

What are the current questions? I might post a few up if there are none.

 

[Zeref]

Re: HSC Physics Marathon 2014

What are the current questions? I might post a few up if there are none.

I don't there are any atm. gogogo

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

I don't there are any atm. gogogo

Will think of some to put on by tomorrow, or possibly tonight.

 

[omgiloverice]

Re: HSC Physics Marathon 2014

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

^^^Do this for Space.

As for Motors and Generators:

 

[Zeref]

Re: HSC Physics Marathon 2014

iirc a geo-stationary orbit has an altitude of 35800km so just add the Earth's radius on top of that.

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

New Question:

Whilst playing tennis, John serves a ball with a velocity of 45ms-1 at an angle of 6.5degrees below the horizontal. The height of the ball as it leaves his racquet is 2.45m above the ground. The net is 12.2m away and is of height 0.92m.
Determine whether the ball clears the net.

Dont forget about my question! xD

iirc a geo-stationary orbit has an altitude of 35800km so just add the Earth's radius on top of that.

Prove it! Never ever pull numbers out of thin air, even if you know them to be true (unless they are on data sheet ofc)

 

[BERRY]

Re: HSC Physics Marathon 2014

Is this correct for the tennis question?

Ux=45cos6.5
=44.71m/s

Uy=45sin6.5
=5.094m/s

Delta x =Ux.t
12.2 = 44.71 X t
t=0.27286.....

Delta y =Uy.t +1/2ay t^2
= 5.094 X 0.2728696 - 1/2 X 9.8 X 0.2728696^2
=1.02515....m

Therefore, it does clear the net, because the height of the net is only 0.92m

 

[Zeref]

Re: HSC Physics Marathon 2014

Is this correct for the tennis question?

Ux=45cos6.5
=44.71m/s

Uy=45sin6.5
=5.094m/s

Delta x =Ux.t
12.2 = 44.71 X t
t=0.27286.....

Delta y =Uy.t +1/2ay t^2
= 5.094 X 0.2728696 - 1/2 X 9.8 X 0.2728696^2
=1.02515....m

Therefore, it does clear the net, because the height of the net is only 0.92m

Yea I got it too
I tried finding time using the delta y formula with delta y being 2.45m and I ended up getting a weird number. Why didn't that work?

 

[Zeref]

Re: HSC Physics Marathon 2014

Dont forget about my question! xD



Prove it! Never ever pull numbers out of thin air, even if you know them to be true (unless they are on data sheet ofc)

Lol for some reason I didn't see your question. I was always on page 3 and I'm like, when is Fizzy_Cyst going to post his question :L

nek minnit: Prove a geostationary satellite has a period of 24 hours

 

[BERRY]

Re: HSC Physics Marathon 2014

Zeref, that is because when you consider it from the perspective of delta y, you are considering the entire time (the time taken to reach ground) however, when you consider it from the perspective of delta x, you only consider the time taken for the ball to be directly above the net (at exactly 12.2m) that's why if you work it out from delta y, I got a time of 0.357973 sec, which is longer than the 0.27286s time achieved by using delta x. Does that make sense?

And so by using the delta x equation, you can get the time taken to get directly above the net and use this in your calculations to find the delta y at this exact point. Does that make sense?

(At least this is what I think)