timeflies
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HSC Physics Marathon 2013-2015 Archive (2 Viewers)
- Thread starter Magical Kebab
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Fizzy_Cyst
Owner @ Sigma Science + Phys Goat
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 just came up with an ub3r hard relativity question, but am going to save it for BoS Physics Trial xD
Zeref
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nexusbrah
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Re: HSC Physics Marathon 2014
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
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
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nexusbrah
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Fizzy_Cyst
Owner @ Sigma Science + Phys Goat
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!Ain't got nothin' on me
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nexusbrah
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Re: HSC Physics Marathon 2014
It doesn't say those details in the booklet?? :3Ur 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!
Fizzy_Cyst
Owner @ Sigma Science + Phys Goat
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
nexusbrah
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Re: HSC Physics Marathon 2014
all your fault </3 Thanks though ill edit the question and work it out for myself as well
OH HAHAHA I DIDNT EVEN NOTICE!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
all your fault </3 Thanks though ill edit the question and work it out for myself as well
nexusbrah
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Re: HSC Physics Marathon 2014
Have to wait for Fizzy_Cyst to respond as he has the right answer.
esaitchkay
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Re: HSC Physics Marathon 2014
still no idea.
saw that question two months ago or so and i've spent quite a bit of time on it.
still no idea.
nexusbrah
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Re: HSC Physics Marathon 2014
Ive got two ans haha hopefully fizzy replies soonsaw that question two months ago or so and i've spent quite a bit of time on it.
still no idea.
anomalousdecay
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Re: HSC Physics Marathon 2014
Speaking of which, did anyone understand my MandG question?
Speaking of which, did anyone understand my MandG question?
Fizzy_Cyst
Owner @ Sigma Science + Phys Goat
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
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
Re: HSC Physics Marathon 2014
My logic gets iffy with non-ohmic stuff. Maybe a new 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
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Re: HSC Physics Marathon 2014
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.I don't know how to do it. Seems more like an extracurricular question than a M&G question
anomalousdecay
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Re: HSC Physics Marathon 2014
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.
(5 \ \Omega) \\ \\ V_{drop} = 50 \ V )
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:
(4 \ \Omega ) \\ \\ V_R = 100 \ V )
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.
New Question for Motors and Generators:
 If a resistor of resistance 4 ohms was placed in series with the motor, in such a way that a limited current flow goes to the motor, find the back emf across the motor, when the current is 25A. (3 marks).$ )
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.
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