HSC Physics Question Thread

[kaustubh.patel]

hey guys need some help with 2012 physics MC q19 and 20, i think 19 is D but the answer is C and 20 i just dont know but its answer is B.

[clovvy]

hey guys need some help with 2012 physics MC q19 and 20, i think 19 is D but the answer is C and 20 i just dont know but its answer is B.

for question 19, electric power raises the temperature of the pot. The pot base has a low resistance. P=I^2R. The more rapid the change in flux, the greater the induced emf. This creates a larger current and more rapid heating saucepan. Therefore it is C
for 20, Max Enerhy output=E=hf=hc/λ =(6.626x10^-34x3x10^8)/(9x10^6)=0.138eV. Therefore the band gap should be less than 0.138eV. This makes HgCdTe the best choice so A.

[Mate2425]

Hey guys so with HSC 2012 Q18. how do you get the figures cause i measured the distance from centre of Earth to point X to be 2.5 and Y to be 3.5 cm respectively?
Also Q20 HSC 2014 could someone please provide me a thorough working out, step by step to this question.

Thank you!!

[clovvy]

Hey guys so with HSC 2012 Q18. how do you get the figures cause i measured the distance from centre of Earth to point X to be 2.5 and Y to be 3.5 cm respectively?
Also Q20 HSC 2014 could someone please provide me a thorough working out, step by step to this question.

Thank you!!

Hi again!!
So for Q18 this is how I do it...

So the answer is C
For Q20 of HSC 2014, this is my solution

[Jane20]

Hi can anyone please show me how to work out the graph question 26b HSC 2014 paper 
Thank you 

[clovvy]

Hi can anyone please show me how to work out the graph question 26b HSC 2014 paper 
Thank you

Before drawing the graph you need to find the frequency first:


Mod Edit: Tidied the LaTex, thanks clovvy!

[Pokechimp7]

Hi everyone,

I need some help with this projectile motion question!

An object is launched at a velocity of 20 m/s in a direction of 25 degrees upward from the horizontal.
a) What's the maximum height reached by the object?
b) What's the total flight time?
c) What's the horizontal range?
d) What's the magnitude of the velocity of the object just before it hits the ground?

Any help is greatly appreciated!

[S200]

Hi everyone,

I need some help with this projectile motion question!

An object is launched at a velocity of 20 m/s in a direction of 25 degrees upward from the horizontal.
a) What's the maximum height reached by the object?
b) What's the total flight time?
c) What's the horizontal range?
d) What's the magnitude of the velocity of the object just before it hits the ground?

Any help is greatly appreciated!

a.) \(v^2=u^2+2as\), where \(v=0\) and \(u=20\sin{25}\)

b.) \(2\times [v=u+at]\), same conditions as above.

c.) Answer from b, \(\times 20\cos{25}\)

d.) Same as launch velocity. \(20ms^{-1}\) at 25 degrees to the negative horizontal.

Image

[Jane20]

Before drawing the graph you need to find the frequency first:


Mod Edit: Tidied the LaTex, thanks clovvy!

Thanks Clovvy !

[not a mystery mark]

Howdyy, I have another question. The teacher today showed me her answer but I couldn't really understand the reasoning behind it. You guys are smart so thank you!

A 600 kg car is merging onto a major highway via a curve banked at 7 degrees from the horizontal.
The radius of the curve is 240 m.
 - What is the maximum safe speed for this car around the curve? -

[blasonduo]

Howdyy, I have another question. The teacher today showed me her answer but I couldn't really understand the reasoning behind it. You guys are smart so thank you!

A 600 kg car is merging onto a major highway via a curve banked at 7 degrees from the horizontal.
The radius of the curve is 240 m.
 - What is the maximum safe speed for this car around the curve? -

Hey! A question from the new syllabus? How exciting!!

An object on a banked track does not use friction, but the velocity of the object to maintain its circular motion. If the velocity is too slow of too fast, the car will either "slip" and fall into the centre of the banked track or go over the top of the banked track. So I'm going to assume that the "safe speed" is the speed for the car to not move up or down the slope.

We know that with any object moving in a circle uniformly, the object travels in a circle at a constant speed due to a force accelerating it inwards, which is called centripetal force. We also know that the force of gravity acts on this car; I've made this diagram to visually help with this.



As you can see, both the gravitational force and the centripetal force are perpendicular, and the Normal force to any object is always perpendicular to the surface, we so know that the angle is 7 in this.

By then equating Tan (opposite over adjacent), we derive a formula (which does not have mass in it amazingly!)



Subbing that in, I get 16.9938 m/s
= 17m/s.

Let me know if i'm right, because "safe" is a broad term. Hope this helps!

[not a mystery mark]

An object on a banked track does not use friction, but the velocity of the object to maintain its circular motion. If the velocity is too slow of too fast, the car will either "slip" and fall into the centre of the banked track or go over the top of the banked track. So I'm going to assume that the "safe speed" is the speed for the car to not move up or down the slope.

As you can see, both the gravitational force and the centripetal force are perpendicular, and the Normal force to any object is always perpendicular to the surface, we so know that the angle is 7 in this.

By then equating Tan (opposite over adjacent), we derive a formula (which does not have mass in it amazingly!)



Subbing that in, I get 16.9938 m/s
= 17m/s.

Let me know if i'm right, because "safe" is a broad term. Hope this helps!

OH MY GOD YOU'RE MAGIC! Thank youu!! That was the answer she got, but this time I've walked away with understanding! Thank you heaps friend. I shall be back with more new syllabus questions. ahaha

[owidjaja]

Hey guys,
I need help with the question attached. I've asked my teacher but I still didn't understand why the answer is C.

[jasn9776]

Hey guys,
I need help with the question attached. I've asked my teacher but I still didn't understand why the answer is C.

ehh its hard to explain but basically, the magnetic field is going into the page from the diagram of the two people.
Using lenzs law you want the current to go from P to Q. So at A, it will generate a current to oppose the change in flux, i.e. a current from Q to P. At C, the current will generate a force upwards. with right hand palm rule the current will go from P to Q.

[radnan11]

Quick question: are we allowed to shorten low earth orbits as LEO in our answers. Thanks