Re: EXAM!!!!!!!!!!!!

Admin · Admin on Mon May 25, 2009 6:55 pm

2008 28 a Irradiance = power / area power is 1mW per 20 mV from question and voltage = 40 mV, area 8.0 x 10^-5m
So I = 2 x 10^-3 / 8 X10^-5 = 25 Wm^-2
2008 29 b hf = hfo + Ek, hf = 6.63 x 10^-34 x 6.1 x 10^14, Ek = 6 x 10^-20
hfo = 40.4 x 10^-20 - 6 x 10^-20
hfo = 34.4 x 10^-20 = 3.44 x 10^-19J

Admin · Admin on Mon May 25, 2009 6:57 pm

Suzy
2007 Q25 b ii yes you do it the same way would you like a worked solution?

Admin · Admin on Mon May 25, 2009 7:00 pm

Suzy the question that you are getting the wrong angle is not 2005 Q27 b ii, if you tell me which one I can give you a solution.

Connor Einstein · Connor Einstein on Mon May 25, 2009 7:20 pm

in 2008 30 b i the the equvilant dose just 0.03 because the weighting factor is 1 so it would be 1*the average absorbed dose

Suzy · Suzy on Mon May 25, 2009 7:50 pm

Okay and sorry its 2005 Q28 b ii

I think it's got something to do with the wavelength of 650nm.
When I make it 650mm it works out to be 23 degrees but when its nanometers it gives a ridiculous number!

Admin · Admin on Mon May 25, 2009 8:16 pm

Right Suzy here goes
2005 Q28 bii
n x wavelength = d sin o; d = 1 / 300000 (300 lines in a milimeter is 300000 lines per meter this might be your problem), n = 2, wavelength = 650 x 10^-9m
Put it into equation 1.35 x 10^-6 = 3.33 x 10^-6 sin o;
therefore sin o = 1.35 / 3.33
o = 23.
2005 Q27 b ii
Voltage at Q = Vs x 20 / ( 100 + 20); Vs = 12 so we get 12 x 20 / 120 = 2V
Vout = (Rf/R1) x (P-Q)
2 = (1000/500) x (P-2)
2 = 2 x (P-2)
1 = P - 2
P = 3V
Voltage at P = Vs x Rt /(75 + Rt)
3 = 12Rt / (75 + Rt); 3 x (75 + Rt) = 12Rt; 225 + 3Rt = 12 Rt
225 = 9Rt; Rt = 225 / 9 = 25 ohms Smile

Admin · Admin on Mon May 25, 2009 8:18 pm

Connor, don't forget the micro sign but yes for 30 b i you got the right idea.

Connor Einstein · Connor Einstein on Mon May 25, 2009 8:27 pm

its been a record breaking night for the forum !!! most things seem to be going right for me now which is good so thre might be a little sunny

at the end. off to bed now see you tommorow !!!

Suzy · Suzy on Mon May 25, 2009 10:04 pm

Aaaaawwwww!!! Yeah I'm an idiot - had d=0.3 instead - right get it now!

And okay, so it's like a voltage divider I understand Smile

unbelievably Cool

Right okay thanks - I think that's everything for now, too late if theres anything else, see you later

S.McGrath · S.McGrath on Mon May 25, 2009 10:41 pm

2008
Q4?

S.McGrath · S.McGrath on Mon May 25, 2009 10:43 pm

2008
Q 12, 13, 14

Admin · Admin on Mon May 25, 2009 10:52 pm

2008 Q4
The max heat energy converted is the Ek lost so find the Ek lost in braking. In otherwords find the kinetic energy of the bike.
Ek = 1/2 mv^2 = 0.5 x 125 x 16^2
Ek = 32000J answer C.
2008 Q12
Inverting mode op amp with a gain of 2 So Vo = -0.2 x 2 = -0.4
Answer B
2008 Q13
Interference is always proof of waves. Answer E
2008 Q14
First maxima so path difference = 1 x lamda = lambda answer C