HSC Physics Question Thread

[jamonwindeyer]

I sort of assumed this was part of an option that I didn't do, where you do proper three-dimensional-like gravity wells? Maybe Astrophysics? In any case, none of us did that topic, and I was definitely sort of making stuff up.

Oh maybe? It says "either side," so I'm thinking a two dimensional thing? Is that right Katniss?

[RuiAce]

I sort of assumed this was part of an option that I didn't do, where you do proper three-dimensional-like gravity wells? Maybe Astrophysics? In any case, none of us did that topic, and I was definitely sort of making stuff up.

I did astro and it doesn't look like astro

[katnisschung]

Thanks Jake and Jamon
yes its a 2 dimensional graph
weirdly its part of space i just started my hsc

ill try posting a lower quality photo becos the other one was too large

[katnisschung]

this is the graph i am referring to

yuna

[jakesilove]

this is the graph i am referring to

yuna

Basically, whilst it's sort of good to understand this, it's far outside the curriculum! So don't worry too much

[FallonXay]

Hi!

How would you solve this graph question? (The answer is C)

Thanks 

[FallonXay]

Oh, sorry,  and this question (The answer is A) 

Thank you!!!

[jamonwindeyer]

Hi!

How would you solve this graph question? (The answer is C)

Thanks 

Oh this is a tough one! Okay, so remember Faraday's Law:



This question requires applying the law above in reverse. Or in words, the induced EMF is proportional to the rate of change of voltage.

This question is made easier if you do 2U and have a basic understanding of Calculus, but think of it this way. The EMF will exist as long as the voltage is either increasing or decreasing, and it will be zero when the voltage flattens out. Further, when the voltage is increasing, the EMF will be negative. When the voltage is decreasing, it will be positive. This is due to the negative sign in the above formula (think Lenz's Law)

This yields C as our answer

[jamonwindeyer]

Oh, sorry,  and this question (The answer is A) 

Thank you!!!

And this one we can answer with a bit of formula. The centripetal acceleration:



However, we can obtain an expression for the velocity:



Combine THAT with a rearranged version of Kepler's Law:
Edit by Rui: Typo spotted



So that yields \(v=\frac{GM}{R}\), and we can substitute that back:



We can substitute our varying values of R (remember that the radius of the earth must be taken into account) to obtain our numerical answers. Or, we can just notice that a higher orbital radius will lower our centripetal acceleration, and so A must be true by default (indeed, this second approach is more time efficient, and may not even require all the working above)

[FallonXay]

Oh this is a tough one! Okay, so remember Faraday's Law:



This question requires applying the law above in reverse. Or in words, the induced EMF is proportional to the rate of change of voltage.

This question is made easier if you do 2U and have a basic understanding of Calculus, but think of it this way. The EMF will exist as long as the voltage is either increasing or decreasing, and it will be zero when the voltage flattens out. Further, when the voltage is increasing, the EMF will be negative. When the voltage is decreasing, it will be positive. This is due to the negative sign in the above formula (think Lenz's Law)

This yields C as our answer

And this one we can answer with a bit of formula. The centripetal acceleration:



However, we can obtain an expression for the velocity:



Combine THAT with a rearranged version of Lenz's Law:



So that yields \(v=\frac{GM}{R}\), and we can substitute that back:



We can substitute our varying values of R (remember that the radius of the earth must be taken into account) to obtain our numerical answers. Or, we can just notice that a higher orbital radius will lower our centripetal acceleration, and so A must be true by default (indeed, this second approach is more time efficient, and may not even require all the working above)

Awesome!!! Thanks a ton!!! 

[teapancakes08]

"A parachutist descending at a constant 4.9m/s dropped is keys when he was 98.0m above the ground. Calculate the time it took for the keys to fall to the ground."

I probably should be able to know how to answer this question but I keep getting an unknown when using formula "s = ut + 1/2at^2"...is there a better way to solve it or did I probably just substitute in the wrong values?

[jakesilove]

"A parachutist descending at a constant 4.9m/s dropped is keys when he was 98.0m above the ground. Calculate the time it took for the keys to fall to the ground."

I probably should be able to know how to answer this question but I keep getting an unknown when using formula "s = ut + 1/2at^2"...is there a better way to solve it or did I probably just substitute in the wrong values?

Hey! I think there may be a better way to do this. Using



We get




So, using





Beautiful! When a method looks too difficult, try something else

Jake

[teapancakes08]

Hey! I think there may be a better way to do this. Using



We get




So, using





Beautiful! When a method looks too difficult, try something else

Jake

Oh, thank you so much! 

This is kind of redundant, but I tried using the other methods after I posted this question and got an answer was a few values off...and then found out I forgot to sub something in after seeing this. Any tips on how to avoid doing silly mistakes like that?

[RuiAce]

Oh, thank you so much! 

This is kind of redundant, but I tried using the other methods after I posted this question and got an answer was a few values off...and then found out I forgot to sub something in after seeing this. Any tips on how to avoid doing silly mistakes like that?

Any extra bits of information you find out you highlight.

As in, over the black print they gave you. Highlight any bit of information that you suspect may be useful. Or even go that extra mile and use highlighting for suspicions and a tick for actual usage of it.

You can also choose to make use of the side of the paper in writing any things you need to sub into the formulae. Some people use this because the markers may or may not care about what's to the left of the lines they give you

Also as obvious as it seems break down any formulas you use. Make sure that what you used makes sense. With practice you should be able to see when things go missing.

[jamonwindeyer]

Oh, thank you so much! 

This is kind of redundant, but I tried using the other methods after I posted this question and got an answer was a few values off...and then found out I forgot to sub something in after seeing this. Any tips on how to avoid doing silly mistakes like that?

Lots and lots of practice, so that when you are doing it you flash back to your earlier work and go, "Yo, this seems a bit off." It's all about experience! You might also want to check out this article!