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Physics: Post your doubts here!

not.maria

not.maria

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Wolfgangs said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
Q15 and Q18., please explain
Click to expand...
for q15
if the depths of water are equal by the end then the gain in height of water in the second tank should be h/2.
since the tanks are identical each tank would have equal volumes so the mass of water that has moved to the other tank should be m/2. since mass is directly proportional to volume provided density is constant. And density IS constant since its water.
Now PE=( m/2)(h/2)g
=mgh/4

Ive posted a query about q18 myself above .didnt get itmyself:D
 
Ahmed H. Al-Neel

Ahmed H. Al-Neel

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not.maria said:
for Q4 use trial and error method until the units on both sides of the equation are equal
L and a has units (m )while T has unit (s)
by using option B we get units on the right hand side equal to the left hand side
View attachment 44333

for q29
For a stationary wave in P there should be a node at the closed end and an antinode at the open end.For a wave to be formed in Q there should be an antinode at both ends.
length of P=35 cm
wavelength of given wave = 20 cm
so 35/20 = 1.75 lambda can fit in P. if you draw a wave 1.75 lambda long starting with a node you will find that it ends with an antinode so this means that a stationary wave would be formed in P
for Q 50/20 = 2.5 lambda would fit in it.
start by drawing an antinode at one open end and continue until you draw 2.5 lambda .u will find that an antinode also forms at the other end.This satisfies the condition for a stationary wave .hence you will get a stationary wave in Q as well
Click to expand...
Thank you sooooo sooo much that helped me aloot i swear to god the way u explain
stuff is waaay better than my teachers!
 
Thought blocker

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not.maria said:
Please can anyone explain:
q8 and q 21
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf

And also can anyone kindly help me out in this?
Click to expand...
8)
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37

21)
Oh,
For these type of questions...i strongly suggest u to assume numerical values yourself..according to question..like if it says P has diameter twice than that of Q..than assume diameter of q=2cm...and therefore P=4cm..these assumptions will make it easy for u, if you still have a doubt, don't hesitate to ask.

18)
There are range of speeds of moelcules.
Some move with slow speed and some move with high speed.
Only some molecules which have higher kinetic energy will move an escape liquid.
And we dont say that solid and liquid have molecular potential enrgy as attractive forces
Hence it is C :)

19)
sugar is a polymer of monossacrides (glucose) and nylon is a condensation polymer so C

20)
total pressure = pressure at surface + pressure due to liquid
so P = ρgh --> 1030 * 9.8 * 80 = 807520
Now add it to 100000 that equals approx of 900000 hence D :)
 
not.maria

not.maria

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Thought blocker said:
8)
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37

21)
Oh,
For these type of questions...i strongly suggest u to assume numerical values yourself..according to question..like if it says P has diameter twice than that of Q..than assume diameter of q=2cm...and therefore P=4cm..these assumptions will make it easy for u, if you still have a doubt, don't hesitate to ask.

18)
There are range of speeds of moelcules.
Some move with slow speed and some move with high speed.
Only some molecules which have higher kinetic energy will move an escape liquid.
And we dont say that solid and liquid have molecular potential enrgy as attractive forces
Hence it is C :)

19)
sugar is a polymer of monossacrides (glucose) and nylon is a condensation polymer so C

20)
total pressure = pressure at surface + pressure due to liquid
so P = ρgh --> 1030 * 9.8 * 80 = 807520
Now add it to 100000 that equals approx of 900000 hence D :)
Click to expand...
Thanks alot! this is helping me soo much:)

however q18 19 20 were for this paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
:)
 
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_Ahmad said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf

Q11,23,25,27,40 PLEASE HELP
Click to expand...
11)
Attached at the end.

23)
We are provided with V = 8 m/s, Max speed = 2πaf, mass = 2 * 10 ^ (-3)
Maximum K.E = 0.5 * m * (max speed)^2
Max speed = 2 * π * 2 * 8/50 -- > We got a from graph, and f from graph itself. a = 2 and f = v / lambda = 8 / 50 (50 from graph)
so K.E = 4 mJ

25)
d sin theta = n lambda
d sin 45 = 3 lambda
d o.7 = 3 lambda
now we are asked to find highest order grating, it means highest value of sin theta, and we know sin 90 has higest value of 1
now we got other equation
d = n lambda
now divide 1 from 2
0.7 = 3 / n
n = 4

27)
D, basic concept yaar.

40)
P = vm
it means p is proportional to m
so alpha has highest mass of 4 so A is the answer

I am in hurry, so gave overview, if you have doubts, surely ask 'em :)
 

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immie.rose said:
Please help!
View attachment 44359
Answer : D
Click to expand...

Since both wires are made of steel, they have the same Young's Modulus value (which is a measure of how easy it is to stretch or compress the material, and is a property of the material itself, not a sample - so, any two pieces of the same material will have the same Young's Modulus).

The formula for Young's Modulus is given by

Y = Fl/Ax = Fl/Ae

Where Y = Young's Modulus,
F = Magnitude of Force stretching or compressing the sample,
l = Length of Sample when no force is applied on it,
A = Cross Sectional Area perpendicular to the Force and the length of the sample, and
x = e = change in length of the sample in the direction of the stretching/compressing force.

So, we know that the extension is the same for both wires (that is what they have given in the question) and the Young's Modulus is the same for both wires - this means that the product eY (or xY) is the same for both samples. Rearranging the formula, we get

eY = xY = Fl/A

So the product Fl/A is the same for both wires.

Let's take P. We don't know what the stretching force is, so we'll simply write it as F(p). We know the length of P is l and that the cross sectional area is A. Therefore, we can write the product as

Fl/A = F(p)l/A

Now let's move on to wire Q. Again, we don't know what the stretching force is, so let's call it F(q). The length of Q is 2l and the cross sectional area is A/2. So, we can write the product as

Fl/A = F(q) * (2l)/(A/2) = 4F(q)l/A

Equating these, we get

F(p)l/A = 4F(q)l/A

Multiplying both sides by A and dividing by l, we get

F(p) = 4F(q)

So that F(p)/F(q) = 4 = 4/1

One last thing we can say is that the tension in the wire is equal to the force stretching the wire (by Newton's Third Law, I guess - i'm not too sure about this part, but suppose you pulling the wire at one end - your muscles are pulling your fingers, and the tension in the rope is pulling them in the opposite direction. The stretching force, in that case, as long as the wire is stationary, must be equal to the tension force. That's my reasoning here, but i'm not too sure about it).

Therefore, F(p)/F(q) = (Tension in P)/(Tension in Q) = 4/1 = D.

Hope this helped!
Good Luck for all your exams!
 
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