Physics P2 Help Required

[intel1993]

q5 a I) as we all knoe wat circumference of the circle is

so no doubt ans is 2 pi r n...........

ii) w.D= Force into distance

so w.d = f * 2pi r n

b) torque= 2 f r....
or u can say fr+fr ...........which is equal to 2fr.........

Total workdone = Forces * dstance

so 2F * 2pi rn

as 2FR=t

so substitute T in place of 2fr........

it wil becum 2pi nT..................(shown)

c)
in this just substitute the values in this : 2pi nT

2 pi (2400)(470)
= 7.09* 10(power 6)

now [power = workdone/tyme
7.09 into 10*6/60
1.2 into 10*5 Watts.....................

 

[MAVtKnmJ]

Got it thanks!

 

[MAVtKnmJ]

Oct/Nov 2004 Q 1 part b (ii) How did we get 8% here? Show working Please

May/June 2006, Q1 part c (ii) I got the speed to be 38320, which is correct but how does this justify our assumption?

May/June 2006, Q2 part b and c (iii)

Oct / Nov 2007 Q1 part b. Please show how you got the answer, especially the uncertainty.

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 ???

Thanks!

 

[MAVtKnmJ]

Anyone?

 

[intel1993]

OCT NOV 2007: Q1 B)

v= PI r l
r= {15/(PI * 20)}*1/2 (UNDER ROOT)
R = 0.489

NOW PERCENTAGE UNCERTANITY......

PERCENTAGE UNCERTANITY OF R / VALUE OF R = 1/2(PERCENTAGE UNCERTANITY OF VOLUME/ VLOUME + PERCENTAGE UNCERTANITY OF LENGTH/LENGTH)

WE USE 1/2 BECAUSE WHEN THERE WE R FINDING UNDER ROOT R (WHEN WE FIND SQUARE WE MULTILPLY IT BY 2)

NOW PUT VALUES IN EQUIATION..

X /0.489 = 1/2( 0.5/15 + 0.1/20)
X= 0.009

 

[intel1993]

M J 2006 Q2 C) iii)

to do it easily we suppose ac =1.........

now clockwise moment = anticlockwise moment..
w * 2/3(1) = T sin beta *(1)
2W= 3 T sin beta ans....

 

[beacon_of_light]

Oct/Nov 2004 Q 1 part b (ii) How did we get 8% here? Show working Please

May/June 2006, Q1 part c (ii) I got the speed to be 38320, which is correct but how does this justify our assumption?

May/June 2006, Q2 part b and c (iii)

Oct / Nov 2007 Q1 part b. Please show how you got the answer, especially the uncertainty.

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 ???

Thanks!

May/June2006 Q1 part (c)... (the easier method)...

The speed that you calculated is wrong. You need to calculate it with the third equation of motion that is v2 = u2 + 2as
Speed will be 9.4 m/s.....Move on to the next part!

Calculate the resistive force by the equation given in part (b)...F = crv...the RESISTIVE force is 3.6 x 10 ^ -5......
Considering 15g calculate the downward force on the ball that is

(15/1000) x 9.81 = 0.15N

So, 0.15N is the weight of the ball while 3.6 x 10 ^-5 N is the upward force..the question says that air resistance is negligible...so it is proved...since 0.000036N is much much less than 0.15N......

May june 2006 part (b)

Since we know for a body to be in eqilibrium, one of the conditions is that the resultant moment must be zero. The lines of action of W and T pass through the same point P...so it is necessary that the line of action of F must also pass through P so that resultant moment of all the forces is zero...as the moment arm of all the three forces from P is zero.....

Part (c)iii

F 's moment about A would be zero sice its line of action passes through A......the horizontal component of T would also have zero moment about A as its line of action also passes through A...now make the equation...

W x AC = T sinB x AB substitue (AC = 2/3 AB)

W x 2/3 AB = TsinB x AB

so solving out the above equation gives.......

2W = 3TsinB

 

[beacon_of_light]

Oct/Nov 2004 Q 1 part b (ii) How did we get 8% here? Show working Please

May/June 2006, Q1 part c (ii) I got the speed to be 38320, which is correct but how does this justify our assumption?

May/June 2006, Q2 part b and c (iii)

Oct / Nov 2007 Q1 part b. Please show how you got the answer, especially the uncertainty.

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 ???

Thanks!

Nov 04..Q1 part B(ii)

Area of cross section = pi x r^2(or the area of a circle)..... you can substitute r = d/2 in the area equation to get another equation that is (pi x d^2)/4.....so

elimination the constants, pi and 4, consider d^2.....I hope you know the rule that when a variable is raised to any power then the uncertainity (denoted here by $)......is mutiplied by the power...that is

( $D/D ) x 2 x 100%

u get ur answer...

 

[MAVtKnmJ]

Nov 04..Q1 part B(ii)

Area of cross section = pi x r^2(or the area of a circle)..... you can substitute r = d/2 in the area equation to get another equation that is (pi x d^2)/4.....so

elimination the constants, pi and 4, consider d^2.....I hope you know the rule that when a variable is raised to any power then the uncertainity (denoted here by $)......is mutiplied by the power...that is

( $D/D ) x 2 x 100%

u get ur answer...

Yeah I did it this way, but what I did was when I reduced the diameter by half (as d/2 = radius) I also reduced the error by half ... so 0.01 mm, but guess I was wrong.
What I don't understand this that when the diameter has been halved then why didn't we half the error as well?

 

[beacon_of_light]

@MAVTKnmJ

you're right...halfing the diameter also requires halving the uncertanity...so again the answer we get is 8%...check that again...

 

[MAVtKnmJ]

I'm still so confused about this question! I've tried it many times but I always make some mistake.

Firstly please tell me, for these kind of questions, what's best, to leave it in diameter form or to change it into radius?

Next can you kindly solve this question again with details about each step? I can't understand what you did with the constant 4, although I do know what you did with pi and the square root.

Thanks!

 

[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!

 

[MAVtKnmJ]

Also, what's the difference between interference and diffraction?

 

[MAVtKnmJ]

I just don't know what the issue is with calculating the uncertainties in the measurement questions! I always get it wrong!

For example check out Oct/Nov 2007 Q1 (b)

Here is my working, please tell me where I am making a mistake!

Calculating Radius

V = pi R^2 L

15 = pi (R^2) (20)

R = 0.489 cm

Calculating the Uncertainty In Radius

This is the formula I use

change in Volume / Volume = 2 (change in Radius / Radius) + (change in Length / Length)

0.5 / 15 = 2 (change in Radius / 0.489) + (0.1/20)

[0.5/15] / 2 = (change in Radius / 0.489) + (0.1/20)

0.0667 - 0.005 = (change in Radius / 0.489)

change in radius = 0.0617 * 0.489

change in radius = 0.0302 cm

So uncertainty +- 0.0302 but that's not the correct answer.

Suggestions?

 

[beacon_of_light]

hey ur misunderstanding...the change in radius...thats actually uncertanitity in radius/radius do it this way...ur calculating change in radius...which is wrong....the correct formula is....

uncert.in V/V = 2(uncert in R/R) + uncert. in L/L

nw u'll get the right ans..

 

[MAVtKnmJ]

I am still confused about what's the difference between "Change" and "Uncertainty" because as far as I know they are the same things ...

Can you please solve this and show me your steps ...

Thanks!

 

[Amanina]

@ 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

 

[intel1993]

interference means meeting of two waves at a point.....................

fiffracyion means spreading of waves when passed through any slit or gap........

 

[ashhadazam]

There is a difference in calculating uncertainties.. here we are calculating the uncertainty in measurement of radius from length and volume not the uncertainty of volume frm radius and length.. so the working u did doesnt fit in this one !

 

[intel1993]

M/j 2006 q5) c:
as we have derived the formula W= 1/2 k(x2^2 - X1^2)
so now put the values.....
as the string r already extended by 4.5 cm.............
so when trolley is koved 1.5 cm either on any side.........
so length of one struing wud be now (4.5+1.5= 6) n one wud be (4.5-1.5= 3)

so now...........

W= 1/2 * 16(100) * ( o.o6^2 - 0.045^2) + 1/2 * 16(100) * ( 0.03^2 - 0.045^2)

as the workdone is done by both of the strings so wiil be added to find out the total workdone.........

so Workdone wud cum out to be 0.36 J....

now speed .......

E=1/2 mv^2
0.36*2/ (850/1000)= v^2

solve it for v.................