National 5 Physics 2015-2016

Here I did the questions for you

I need help with Sqa 2014 paper 2 qu2 please.

When you say the angle of water needs to be smaller than the angle or air. Is the angle you are talking about measured from the normal line to the ray ?

Wow Wow Wow, you never fail to impress me, haha

Thank you very much !

I got 65% on the specimen paper. What is that as a grade?

The notional grade boundaries are:
A 70% +
B 60% - 69%
C 50% - 59%
D 45% - 49%
No Award 0% - 44%

Note that each exam is different and so grade boundaries can be adjusted accordingly, just like last years higher maths exam, and probably this year nat 5 exam, the grade boundary was adjusted to 35% for a C.
So to answer your question, 65% is a B.

Urghh. Really wanted to get an A but oh well I'm happy with B. Guess i'll do more questions and go over the topics again Thank you xx

Maybe it has something to do with doing papers at 2am in the morning??? working too hard won't help a lot.

Does anyone know how to do Question 3 on the 2015 past paper. It's so obscure...

Does anyone know how to do Question 3 on the 2015 past paper. It's so obscure...

Does anyone know how to do Question 3 on the 2015 past paper. It's so obscure...

I'm guessing it's section two Question 3

3a. (i) - As you can see from the diagram, position Y is more closer (in length/distance) to the steel sample than Position X therefore Position Y will be transmitted in less time than Position X. Therefore you can easily identify that the time taken between the pulse being transmitted and received at Position X is 15microseconds

(ii) - You want to calculate the length of the steel sample i.e it's width at position X (i don't know if I'm making sense but just let me know if i'm confusing you). You are given the speed of ultrasound in steel which is 5200ms-1. And you have also concluded from the previous question that the time for Position X to be transmitted and received is 15microseconds. Therefore you can use d=vt to calculate the distance of the steel sample.

Note: the time is in microseconds so do d=5200 x 15x10-6. This equals to 0.078m.

Then to calculate the thickness you divide 0.078 by 2 to get 0.039m.

b. So from the graphs and diagrams you have concluded that Position X has a longer distance than Position Y. However position Z is in between them. So you would just draw a line that is between Position X and Position Y to get the mark.

c. (i) - Simply do f=1/T. i.e f= 1/4.0x10-6 (because 4.0 is in microseconds) and you should get the answer 2.5x10^5

(ii) - You would use v=fWavelength
wavelength=v/f
= 5200/2.5x10^5
= 0.021m

d - The speed of ultrasound in brass will be less than steel because in brass it takes longer for it to travel the same thickness.

Hope this helps

JHEEEEEEEEEEEEEEEEEEEZ!

Why are you only getting 65% in the past paper?

You should be getting A's, you seem very well prepared

Smart.

Is this correct ....??

Volume increases = Pressure decreases
Temperature Increases = Pressure increases
Tenperature increases = Volume increases

How do you change a volume from litres to m^3 ?