Mr. Study's 3 and 4 Question.

[Mr. Study]

EDIT:

Could someone check my calculations?

Question: A star has a mass of 3.0 X 10³⁰ kg and a radius of 10 km.
Find gravitational field strength.

X X
X

The answer is

And how would I work out the gravitational field strength, 5000km away from this star?

Thank you very much. 

[rife168]

EDIT:

Could someone check my calculations?

Question: A star has a mass of 3.0 X 10³⁰ kg and a radius of 10 km.
Find gravitational field strength.

X X
X

The answer is

And how would I work out the gravitational field strength, 5000km away from this star?

Thank you very much. 

Remember to convert the radius to metres!

5000km away from the surface? then you would use 5000+10 (km) as the radius. And you would remember to convert to metres of course!
(If you meant 5000km from the centre of mass of the star, then using 5000km would be fine)

Also, to do times in LaTex, use \times and a dot is \cdot
i.e.

[Mr. Study]

Thanks for that.

Converting, my biggest downfall. Haha. :

[Mr. Study]

Just this electricity question:
                                   The voltage is 6.0V, Resistance is 100 ohms and the diode is an LED. How would I find the potential difference?
     _________
__|__             |
  ___            _|_
    |              |    | 
    |              |__|
    |               _\/_
    |_________|

Thanks

[appianway]

LEDs should have a constant potential difference across them when in operation. The remainder of the voltage drop must take place across the resistance in the circuit, which then dictates the current in the circuit.

[Mr. Study]

I see! Thank you for that appianway. I probably should revise my definitions on what an LED is.

Just this quick question for today.

I have kinetic energy to be 225J and the total work done to be 1500J. The question asks for the amount of energy converted into heat.

I thought the answer would be 225J, as Kinetic Energy is movement energy which, I think, can be attributed to heat. :S.

However the answer is 1275J, which is the total work - kinetic energy. Why is this the case?

Thank you.

[appianway]

Heat which can be attributed to kinetic energy is attributed to the kinetic energy of the random motion of particles. What you're looking at here is the kinetic energy due to the motion of what's probably a reasonably large object, and it's also not moving randomly. So using conservation of energy, this kinetic energy + heat energy (which we'll treat differently to kinetic energy here, even though it's due to kinetic energy of different particles) = work done.

[Mr. Study]

Just this quick question.

For acceleration due to gravity, Does VCAA accept it as 9.8 ms-2 or 10 ms-2? I've been doing it as 9.8 ms-2 but after doing the itute exam (Motions Section Only), the answers, when compared to mine, are marginally different. This is due to myself using 9.8 ms-2 whereas they're using 10 ms-2.

I would look at the VCAA Assessor Reports but I don't want to spoil anything. :O

Thanks!

[pi]

In physics it's 10m/s^2

[Mr. Study]

:O.

I've been using 9.8 ms-2 since Year 11. My teacher said either didn't matter... -__-. Oh well, Never too late to change. Haha.

I tried searching around for it but I couldn't get a definitive answer through a very rough skim of the VCAA assessor reports.

Thanks for the answer.

[Lasercookie]

Just look at the actual exams, they all say (2004 Pilot exam and onwards) at the start of the motion section:

"You should take the value of g to be 10 m/s^2"

[Mr. Study]

Thats embarrassing. Read your post, Looked at the exam. :S

I read the front page and skipped the Section A Instructions before posting my question. :O. I probably should be listening when people say, 'Read all intructions'.

Thanks for that laseredd.

EDIT: Could someone help me with this question. Pictured below.

I had to read the answers but I still don't quite understand why.

How is R₁ and R₃ parallel? I tried turning the page but that doesn't help.  . I am not too good at re-drawing this circuit due to that line connecting the resistors.

I thought that R₁ and R₂ are parallel, R₃ and R₄ are parallel and once we work out total resistance for each parallel resistor, we could just add the two answers to get total resistance, since it'll be in series, for the circuit.

I got the answer wrong and I thought the line connecting the lines wouldn't do to much.

NVM, I got it right! I thought the resistors were in a different order. Sorry!

[Mr. Study]

This question is from the 2008 itute exam.

I decided that since it was asking at the highest point, it was asking for the vertical velocity, which would be zero.

However, the answer is 19.1 ms^-1 (This is exactly what the answer said: At the highest point, vertical component of velocity = 0. Horizontal component of velocity is 19.1. Speed is approximately 19.1), which I also calculated but I decided not to use it as I didn't feel the question was asking for the horizontal velocity. (I associate Vertical speed with height and horizontal speed with distance).

Okay, So my question is, from reading that question, What interpretation would you get from it? Is it asking for vertical/horizontal? If it's unanimous that it was asking for horizontal, I'll be  .

Thanks!

[Bhootnike]

No no you were right, to an extent.

its just that you had to do the whole triangle to find resultant/ speed = horizontal + vertical thing, i think.
What i mean by that is, the velocity vector has 2 components, and so you have to use pyhag to find resultant.
 and because the vertical was 0, and horizontal was 19.1, your net is going to 19.1

[Mr. Study]

Oh! I see why you would do the net speed and NOT the individual components of that net speed.

Thanks for that Bhootnike.