HSC Physics Marathon 2013-2015 Archive (1 Viewer)

[timeflies]

Re: HSC Physics Marathon 2014

So 1:8?

Yes that is correct!

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

I just came up with an ub3r hard relativity question, but am going to save it for BoS Physics Trial xD

 

[QZP]

Re: HSC Physics Marathon 2014

I just came up with an ub3r hard relativity question, but am going to save it for BoS Physics Trial xD

I look forward to that.

 

[Zeref]

Re: HSC Physics Marathon 2014

I just came up with an ub3r hard relativity question, but am going to save it for BoS Physics Trial xD

come at me

please dont....i was bluffing

 

[nexusbrah]

Re: HSC Physics Marathon 2014

A car comes to a bridge during a storm and finds out the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side of the road the car is on is 21.3m above the river, while the opposite side is a mere 1.8m above the river. The river itself is a raging torrent 61.0m wide.

a) How fast should the car be traveling at the time it leaves the road in order to just clear the river and land safely on the opposite side?
b) What is the speed of the car just before it lands on the other side?

a)

b)

New Question
A target appears at a distance of 1250m with the centre of the target 1.2m above ground level. If the sniper was firing from the prone position, calculate the angle to which the barrel of the rifle should be raised to hit the centre of the target. The muzzle velocity is 800ms^-1. The riffle is 30 cm above the ground when in the prone position. 3 marks

"Hard" Question

 

[nexusbrah]

Re: HSC Physics Marathon 2014

I just came up with an ub3r hard relativity question, but am going to save it for BoS Physics Trial xD

Ain't got nothin' on me

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

New Question
A target appears at a distance of 1250m with the centre of the target 1.2m above ground level. If the sniper was firing from the prone position, calculate the angle to which the barrel of the rifle should be raised to hit the centre of the target. 3 marks

"Hard" Question

Ur missing the muzzle velocity of 800ms-1 and the fact that when in the prone position, the muzzle is 30cm off the ground

Ain't got nothin' on me

Fighting words, mate!

 

[nexusbrah]

Re: HSC Physics Marathon 2014

Ur missing the muzzle velocity of 800ms-1 and the fact that when in the prone position, the muzzle is 30cm off the ground




Fighting words, mate!

It doesn't say those details in the booklet?? :3

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

It doesn't say those details in the booklet?? :3

Because although its on a new page, under a new heading, its part (d) of the question! All the other info you need is in the main description of the question

 

[nexusbrah]

Re: HSC Physics Marathon 2014

Because although its on a new page, under a new heading, its part (d) of the question! All the other info you need is in the main description of the question

OH HAHAHA I DIDNT EVEN NOTICE!

all your fault </3 Thanks though ill edit the question and work it out for myself as well

 

[nexusbrah]

Re: HSC Physics Marathon 2014

Anyone wanting to answer the question?

 

[hypermax]

Re: HSC Physics Marathon 2014

is the answer 0.0413 degrees

 

[nexusbrah]

Re: HSC Physics Marathon 2014

is the answer 0.0413 degrees

Have to wait for Fizzy_Cyst to respond as he has the right answer.

 

[esaitchkay]

Re: HSC Physics Marathon 2014

Anyone wanting to answer the question?

saw that question two months ago or so and i've spent quite a bit of time on it.
still no idea.

 

[nexusbrah]

Re: HSC Physics Marathon 2014

saw that question two months ago or so and i've spent quite a bit of time on it.
still no idea.

Ive got two ans haha hopefully fizzy replies soon

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

Speaking of which, did anyone understand my MandG question?

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

Answers for the sniper question are 0.59º and 89º45'



I will give you a hint: It ends up as a quadratic in Tan.

If u want the worked answer, I will post up

 

[QZP]

Re: HSC Physics Marathon 2014

Speaking of which, did anyone understand my MandG question?

I don't know how to do it. Seems more like an extracurricular question than a M&G question My logic gets iffy with non-ohmic stuff. Maybe a new question?

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

I don't know how to do it. Seems more like an extracurricular question than a M&G question My logic gets iffy with non-ohmic stuff. Maybe a new question?

Its actually a bit of skills stuff which is found throughout. I will post a solution and think of a new question in the next few days.

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

New Question for Motors and Generators:

a) Voltage drop is equal to the input voltage minus the back emf.

Also, the voltage drop is equal to the required current to keep the motor running.





b) In Series, the current is the same throughout the circuit. So the motor will receive 25 A of current indefinitely under this specific load.

The Voltage however will be split so that both add up to 240 V altogether.

Given that the motor will receive 25 A of current, from the graph this translates to 2 ohms of resistance.

The resistor of 4 ohms will receive 25 A of current through it, and hence will have a potential difference of the following across it:



So now we know that the Voltage across the motor is:



So to calculate the back emf across the motor, we must take find the voltage drop, which is:

{tex] V_{drop} = IR \\ \\ V_{drop} = (25 \ A)(2 \ \Omega) \\ \\ V_{drop} = 50 \ V [/tex]

Notice that the Voltage drop is the same at all times from the graph. This occurs as the normal load undergoes a variable resistance as the motor is not ideal, as characterised through the graph.

So we get that the back emf is:




Will think of something new and post up as soon as I think of something.