Physics: Post your doubts here!

[Thought blocker]

THNX
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf q22

Here you go bro!
use the formula for E
E = F L / Ax
rearrange to get the ratio x / L on one side (change in length / original length)
you'll get is x / L = F / E A : Where - (A = pi r ^2 )
= 20 / 2 * 10^(11) * pi x (2.5 * 10^(-4))^2
= 5.1 * 10^(-4)
multiply this by a 100 to get the percentage
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %
So answer is B.

 

[ZaqZainab]

Take centre of moment 2 b on pivot n resolve
Clockwise moments = Anticlockwise moments
(10 × 100) + (50 × D) = (20 × 60)
1000 + 50D = 1200
50D = 1200 - 1000
D = 200/50
[ D = 4 cm ]

So distance from pivot = 4 cm
Distance from 0 mark = 40 + 4 = 44 cm

where did you get 10 from?
the 10*100??

 

[Thought blocker]

where did you get 10 from?
the 10*100??

LOL! he posted the answer again, I think he might not have got my alert.

 

[Hadi Murtaza]

where did you get 10 from?
the 10*100??

It is da distance of da pivot from da centre of mass

 

[MYLORD]

Here you go bro!
use the formula for E
E = F L / Ax
rearrange to get the ratio x / L on one side (change in length / original length)
you'll get is x / L = F / E A : Where - (A = pi r ^2 )
= 20 / 2 * 10^(11) * pi x (2.5 * 10^(-4))^2
= 5.1 * 10^(-4)
multiply this by a 100 to get the percentage
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %
So answer is B.

one more !! Man i am sooo busted !!
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf q 17

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf

Q16,21,23,24 help!!!

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_11.pdf

Q16,21,23,24 help!!!

16)
The vertical components of both H and W are cancelled by the upwards vertical component of T. Furthermore, T balances out the horizontal component of H too. Thus T has to be the largest of the three. C is the only option.

21)
As Menu answered

23)
F = kx so x = F/k
For A : x = 4/k
For B : x = 3/k
For C : x = 3k
For D : x = 8/3k
So here constant k is either being multiplied or divide, hence we take k as 1 if we would have addition or subtraction included, we take k as zero
Now put 1 instead k .. A = 4, B = 3, C = 3, D = 2.5 so A is the answer

24)
As Menu answered

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_13.pdf

Q4

 

[MYLORD]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf q 11

 

[MYLORD]

it's A because v=w-u
w would be in the same direction and u would be upwards

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf q 11

 

[MYLORD]

thnx

F=ma
10-4=[20/10]xa
a=3ms^-2

 

[ShreeyaBeatz]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
question no 8 and 4

 

[ShreeyaBeatz]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
qno. 8 !

 

[MYLORD]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf q15

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf q15

Power = Force * velocity
Power = (1000 + 1000*10) * 0.5
Power = 11000 * 0.5
Power = 5500 W = 5.5 kW
Answer: B

 

[zem]

Can someone help me with this question?

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
qno. 8 !

using s=ut+0.5gt^2
for distance x,
x=0*t1+0.5gt1^2 (since initial velocity =0)
x=1/2gt1^2 (eqn 1)

for distance x+h,
x+h=0*t2+1/2gt2^2
x+h=1/2gt2^2
making x the subject gives x=1/2gt2^2-h (eqn 2)
since x=x
eqn 2=eqn 1

1/2gt2^2-h=1/2gt1^2
2(1/2gt2^2-h)=gt1^2
gt2^2-2h=gt1^2
gt2^2-gt1^2=2h
g(t2^2-t1^2)=2h
g=2h/(t2^2-t1^2)

 

[Hijab]

Can some one plzzz solve this question... I would really be grateful

 

[_Ahmad]

Can some one plzzz solve this question... I would really be grateful

check the 11th post of the previous page Menu Mendz has already solve it.

 

[sagar65265]

Can someone help me with this question?View attachment 43858

Wow, is this a good question! The trick here is to simply work through all the steps, and see what you get.
First step is in find the extension from the values we know.

The Young's Modulus Formula is

Y = Fl/Ae (where F is the tension, l is the original, unladen length of the sample, e is the extension, and A is the cross section perpendicular to the extension)

So, by multiplying both sides by e, we get

eY = Fl/A From here, suppose we divide both sides by Y, we get

e = Fl/AY which is the extension formula that we need. What we need to do is find out the values of this ratio for both the scale model and the full size version, and divide them to get the final answer.

A very important point to note here is that the question says that all linear dimensions are in a ratio of 1:10; this means that quantities such as length, radius, diameter, height, and so on are 10 times greater in the real version than in the scale model. Quantities such as Area, Volume, and Density are non-linear and so do not follow this ratio.

On reading the question again, there is something crazily interesting there - they have clearly mentioned that the load is cubic. Why? WHY?
It's important, that's why:

Suppose the material has some density ρ and it has a side length a. Then, for the scale model, the mass = ρV = ρa^3.
For the full size model, the same follows - however, the length has increased to 10 * a, so the mass of the load in the full-size crane = ρ(10a)^3 = 1000ρa.
Major importance there!
I'm pretty sure there is no more to be taken in consideration, since the extension of a wire requires you to consider only the dimension of the wire (going to do it), the load (done by above discussion) and the kind of material being stretched (Young's Modulus is the same since the material used is the same. Done!)

So, for the scale model:
i) The force can be written as "F" (let's say). This is also equal to ρa^3, as discussed above.
ii) The length of the cable without any strain applied can be written as "l".
iii) The Cross Sectional Area of the wire should be written as πr^2, where r is the radius of the cable.
iv) Finally, the Young Modulus can just be written as "Y".

Therefore, we can write

(extension of cable on the model crane) = [(ρa^3)gl]/Yπr^2

For the real crane:
i) The force can be written as "1000ρa^3", again as discussed above.
ii) The length of the cable without any strain can be written as "L". We know from the question that this is equal to 10 * l, so the length = "10l".
iii) The Cross Sectional Area of the wire should be written as "πR^2" where R is the radius of the full scale wire. We know this is equal to 10*r, so we can write that
the cross-sectional area of the real cable = π * (10r)^2 = 100πr^2.
iv) Again, the Young Modulus is just Y, since the material of the real cable is the same as the material of the scale model cable.

Therefore, we can write

(extension of cable on the full-size crane) = [(1000ρa^3 ) * 10l]/[100Yπr^2] = [(100ρa^3)gl]/Yπr^2

Dividing those two, we get (extension of cable on the full-size crane)/(extension of cable on the model crane) = 100 = 10^2 = C.

Hope this helped!
Good Luck for all your exams!