OCR B Physics G494 June 11th 2014 Unit 4

Hi,
Can someone help me with question 6b on June 2010,
To work out the internal energy in the tyre, why can you use 3/2NKT and NKT to get the answer.
Which one is better to use , and where does the 3/2 come from . Thanks

Hi,
Can someone help me with question 6b on June 2010,
To work out the internal energy in the tyre, why can you use 3/2NKT and NKT to get the answer.
Which one is better to use , and where does the 3/2 come from . Thanks

Hi guys, I'll give you a chance to test your knowledge by helping me here. The question is 11 on June 2011 (http://www.ocr.org.uk/Images/62268-question-paper-unit-g494-rise-and-fall-of-the-clockwork-universe.pdf)

I can't do part b).

Other than this, I'm finding the papers reasonably easy. How about you guys? How many exams has everyone had/got left? I've done 6 and I've got 5 left.

I'm not the brightest student but I will attempt to give my helping hand.

Bi) You have previously worked out/been told that the resistance is 38000(Ohms) whilst the capacitance is 1000x10^-6(Farrads). The product of these is the time constant (RC). This is constant throughout the table at a value of 38. The time interval (delta)t is 20 seconds and the charge is in the 2nd column. Put these values into the formula given in the heading of the third column and you have your loss of charge. Add this value (Remember it is negative) to the initial charge for that time frame to get the new charge for the next row and so on.

Bii)The model assumes that the charge is constant throughout the time interval with the value that is given at the start of the time interval, in reality it is dropping all the time (Exponential decay). This means that the discharge value (-Q) is too big and therefore causes the discrepancy between the two sets of data as the model removes more charge then it should each time interval.

Sorry if this made absolutely no sense..

Thank you so much! I knew it had to be something simple, but I just couldn't see how to get the Q for the next row. It never occurred to me that it could be something as "complicated" as adding the Q and delta Q from the previous row. Hopefully I won't do something like that tomorrow!

Many thanks again!

Could anyone please help me with the following question. Even after looking at the mark scheme I can't get the correct answer. Question 4(b) on June 2010.

June 2010 - http://www.ocr.org.uk/Images/59861-question-paper-unit-g494-rise-and-fall-of-the-clockwork-universe.pdf

Was just about to answer but it looks like I've been beaten :P also, not to sound patronising or anything, but make sure you read the question properly, I nearly missed the FIFTY years when doing the calculation (because they gave the conversion for a year in seconds and I typed that in), even though the activity after fifty years was the whole point of the question.

First calculate the acceleration of the mass using the equation they have given (a=-50x). Once you've calculated the acceleration plug this value into a suvat equation to find the final velocity. (v=u+at). Once your final velocity has been calculated you must find the average of these values ((v+u)/2) and then using s = ut find your displacement. Now add this value to your initial displacement and you have calculated the total displacement from the equilibrium point!