Physics P2 Help Required

[MAVtKnmJ]

Ah I get it, I have to switch the formula and make it according to the required parameters.

Thanks for the help fellass!

Can someone explain the other questions I asked.

Thanks in advance!

 

[MAVtKnmJ]

@ MAVtKnmJ

maybe u can just directly find uncertainty for radius from this equation

R(square) = V/(L X pi)

then it will be like this 2(uncertainty in R) /R = (uncertainty in V) / V + (uncertainty in L)/L

2 (uncertainty in R) /0.489 = 0.5/15 + 0.1/20
2 (uncertainty in R) = 0.0383 X 0.489
uncertainty in R = 0.0187 / 2 = 0.009

Thanks, in my post when I said "change in a quantity" like Volume, Length, etc I did indeed mean the uncertainty. If you see my post it's almost alike, but I guess I was doing some mistake during the algebric manipulation.

Now since I've made R the subject of the formula as you had advised, I get the right answer

Thanks!

 

[intel1993]

hey can any one of u can post the ms paint diagram of deviations of x-partices in gold foil experiment question..............as asked in n2004 q7 b)

 

[MAVtKnmJ]

I've sent you the PM intel. But here is the image again.

http://i46.tinypic.com/2rc2w5g.jpg

Hope that helps

 

[WellWIshER]

cany any one tell how to find the spped of the wave in mj 2007 question????

 

[Azazel]

S07 Q5 b) ii)

We can see there are 5 anti-nodes in the 39cm region (Each empty space is an antinode, you can draw the stationary wave itself if you're unsure on how many wavelengths are in the 39cm).

One wave has two antinodes, therefore there are 2.5 complete waves in the 39cm region.

2.5 : 39
1 : x

where x is the wavelength.
x = 39/2.5 = 15.6 cm = 0.156m

v = f * lambda
v = fx
v = 2140 * 0.156 = 333.8 m/s

 

[MAVtKnmJ]

Some more problems!

May/June 2004, Q6 (b) - Can you cut the image and use windows mspaint to show me where CC and DD are supposed to be? Or if it's easier link me to a website which has exact information?

May/June 2006, Q5 (c) - Kindly show details about each step.

May/June 2006, Q6 (a) and (c) - Kindly show details about each step. - For this question, the waves produced are in a closed pipe right? The water running from the tap is just a trap to make you think it might be an open pipe right? Secondly in this question how do you find out the shape of the wave, I mean how do you know how many nodes and antinodes it will have? Is it because there were two noises?

Thanks, your a big help man!

Would appreciate insight on the above!

 

[MAVtKnmJ]

Also with the following.

May/June 2009 Q5. Kindly explain and show working. Plus is the path difference in a destructive interference (n-1/2)laimbda or (n+1/2)laimbda ???

 

[Azazel]

Some more problems!

May/June 2004, Q6 (b) - Can you cut the image and use windows mspaint to show me where CC and DD are supposed to be? Or if it's easier link me to a website which has exact information?
..
!

Would appreciate insight on the above!

Sorry for the crap drawing,

I'm using a touchpad.

Basically, the open spaces are where destructive interference is taking place & the points where they cut each other are where constructive interference are taking place.

 

[sizbeauty]

for distructive interference,path difference is(n-0.5lamda)

can sm1 plz help me out wid june08:
Q4(part c)
Q6(part b)don't we just add up the powers???

 

[sizbeauty]

Also with the following.

May/June 2009 Q5. Kindly explain and show working. Plus is the path difference in a destructive interference (n-1/2)laimbda or (n+1/2)laimbda ???

phase difference=( 2pi/ lamda) X path difference

phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi X path difference.
path difference is S2M - S1M
find S2M using Pythagoras = root of 802 + 1002
=128
path difference= 128-100=28
wavelength= 2pi/pi X 28
=56cm
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas

 

[DREAMGIRL]

HEY MAVtKnmJ
YOU KNOW THAT F=CHANGE IN MOMENTUM / TIME NEWTONS 2ND LAW
VELOCITY IS IN DIFFERENT DIRECTIONS SO MOMENTUM WILL DE ADDED.
0.035(3.5--4.5)/ 0.14
=2
OPPOSITE INITIAL DIRECTION

c) MOMENTUM IS ALWAYS CONSERVED IN A CLOSED SYSTEM SO WHATEVER MOMENTUM THE PLATE GAINS IS EQUAL AND OPPOSITE TO THE CHANGE FOR BALL.

 

[Azazel]

for distructive interference,path difference is(n-0.5lamda)

can sm1 plz help me out wid june08:
Q4(part c)
Q6(part b)don't we just add up the powers???

Q4 c)

According to the diver's assumption,

101000 = 1080 * 9.81 * h
h = 9.53m

The error in h, is therefore, 10 - 9.53 = 0.47m
Percentage error= [error in h/h] * 100 = [0.47 / 10] * 100 = 4.7%

Q6 b)

Someone posted a really neat response to this question in another thread, let me quote:

may 08 Q6 b...

Just remember one rule, the current ''likes'' to flow through the path with lowest resistance. here in this diagram, if switch S2 would be closed, current will flow through switch S2 but not through resistance B.

row1: power is zero since no current flows through circuit.

row2:now total resistance will be 38.4 ohms only coz current will not flow through resistance B as an alternative low resistance path by S2 is provided.

power is v^2/R .....(240)^2/ 38.4 ...power is 1.5 ohms.

Solve out other rows in the same way...

 

[Amanina]

May/June 2006, Q5 (c)

both springs lets say A and B have initial length 4.5cm. after moving the trolley, spring A will extent 1.5cm while the spring B will compress 1.5cm.
so, final length spring A is 4.5 + 1.5 = 6 cm
spring B 4.5 - 1.5 = 3 cm

use expression in b
convert every cm to meter

workdone in spring A = 1/2 (1600)(0.06^2 - 0.045^2)
= 1.26J

work done in spring B = 1/2 (1600)(0.03^2 - 0.045^2)
= -0.9J

total workdone = WD in A + WD in B
= 1.26 +(-0.9) = 0.36J

WD = 1/2 mv^2
0.36 = 1/2 (0.85) (v^2)
v= 0.92 ms-1

 

[MAVtKnmJ]

Thanks for the help and detailed information fellas!

 

[MAVtKnmJ]

Kindly have a look, I've drawn the slanted line, is it the correct line for a destructive interference?

 

[Amanina]

MAVtKnmJ, yes it's correct.

 

[Azazel]

I have a question.

W07 P2 4 (c) - I've got the area to be 2.0 x 10^-6 but I don't understand why we're subtracting this from 3.2 x 10^-6.
4 (d) - I don't exactly get the logic behind the mark scheme's answer!

 

[MAVtKnmJ]

for distructive interference,path difference is(n-0.5lamda)

If this is correct then how come in the following post "beacon_of_light" has used " n + 0.5lambda " and yet she got the right answer ???

viewtopic.php?f=26&t=1767#p18457

Thanks!

 

[Azazel]

It's the same thing: try understanding what the equation means. n is the point where displacement is max, if you add half a wavelength to it, you'll be where displacement is zero. If you remove half a wavelength, you'll still be where the displacement is zero.