Post by: Cravecom on December 03, 2009, 10:23:44 pm
I can do part a of question 7, but part b and question 8 i have no clue :S
[IMG]http://img512.imageshack.us/img512/5364/physicsq.png[/img]
Thanks in advance for any help
Post by: /0 on December 03, 2009, 10:37:42 pm
Let the mass of the asteroid be
The force on the asteroid from Alpha is
The force on the asteroid from Beta is
If we have
You can solve for
8.
Let point Y be y metres from Alpha. At this point,
And you can solve for
Post by: Cravecom on December 04, 2009, 08:10:33 pm
Question 8 though seems a bit hard. I understand the method, but solving for Y seems a bit arduous. Would you be expected to solve a complex equation like that in an exam situation?
Post by: /0 on December 04, 2009, 08:38:47 pm
Post by: Cravecom on December 07, 2009, 07:52:56 pm
This time i'm back with a few electronics Q's which i'm struggling to get my head around, and again; any help at all would be greatly appreciated.
(http://img32.imageshack.us/img32/9736/questionsx.png)
And here is Fig 5.13 required for Q. 5
(http://img30.imageshack.us/img30/7579/fig513.png)
Post by: Cataclysmic on December 07, 2009, 08:04:17 pm
50M = 50*10^6
I = 50 * 10^-9
:. Voltage drop at the resistor =
IR = 50*10^-9 * 50*10^6 = 2.5
So the diode takes 6 - the above voltage
= 3.5
B) I forget their proper definitions but on page 160 of the Heinemann text book, it says " A reverse biased photodiode operates in what is called photo conductive mode". I'm not sure about the other one but taking a random guess, maybe not reverse-biased? lol
http://optoelectronics.perkinelmer.com/FAQs/FAQdisplaytable.aspx?FaqId=17
Post by: Cravecom on December 07, 2009, 08:26:30 pm
I was trying to use the Voltage divider formula and getting some answer of like 4V :uglystupid2:
Ahhh and also thanks to you, i have been able to solve Q5. :P
I've just been trying to over complicate things when i should just go back to the good old V = IR :)
Post by: Cataclysmic on December 07, 2009, 08:28:23 pm
b)
R gets the same voltage as the LED because they
are parallel.
T receives 7.5V ( since the whole supply is 10V)
V = I R
7.5 = I (500)
I = 0.015 A ( current at T)
since 0.011 has gone to the LED
0.015-0.011 goes to R.
V = IR ( at the resistor)
2.5/0.004 = R
R = 625
c) temperature relates to the resistance of the thermistor ( the graph), so
find the resistance at point T when the LED has a voltage of 2.
If V = 2 at LED, V = 2 at R.
So 8 V goes through T.
Using V = IR at r,
2 = I (625) (625 from previous question)
I = 0.0032
A current of 4.8*10^-3 goes through the LED. ( given in question)
So I at T is 0.0032 + 0.0048
Using V = I R at T
8 = (0.008)R
R = 1000 = 10 degrees.
(btw: the pictures are blury ;d)
Question 9:
Switch on 2 V, draws 20mA for optimal...
A), for them to get 2V each, Rl needs a voltage
drop of 7V. ( 1st circuit). Since the 3 globes are in parallel, the current would
be 60mA. V = IR, so R = V/I = 7/60*10^-3 = 116.67 ohms
They are in series, so they'll receive the same current but not voltage.
Rl would need to leave 6 V ( so that they get 2 V each )
meaning Rl should take 3 V.(2nd circuit)
They are in series, so I = 20ma
3 = (20*10^-3)R, R = 150 ohms
B) TBH - I wasn't really that sure and if I had to answer it, it would of been a guess.
I intepretted "most light" as "highest power" but since it's telling you that the voltage
at each globe has to be 2V and that it draws 20mA (too little = dim, too much = burn out), I would say
that using P = VI, they are the same. I think you should get someone else to answer this, I suck with explanation stuff :)
C) Looking at the 1st circuit, it requires 2V for each battery
and 60mA in total ( current).
Circuit 2 requires 2V for each globe as well but only 20ma( since they are in series) meaning
the power usage of the 2nd circuit in less. So the 2nd circuit will last longer.
(sorry if I include extra rubbish, I usually just write stuff out even when it's already given).
5)
Looking at the graph, as the intensity goes up, the resistance goes down.
The resistance is 25,000 at 100mW m^-2,
so for the work place to receive an intensity higher than the acceptable amount,
the resistance at the LDR for a specific out-put must be lower than 25,000.
Vout = 4V, so Vout at R = 8V.
V = IR at (R)
8 = I(20,000)
I = 4x10^-4
at the LDR, I = 4*10^-4
V = 4V
R = ?
V = I R
R = V/I = 4/4*10^-4 = 10,000 ohms
This equates to a light intensity greater than the "minimum acceptable level"
b) Really bad question here imo.
I guess it's just reading off the graph at where the resistance is 10,000. ( which
is between 400-600 mW m^-2)