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Physics: Post your doubts here!

Thought blocker

Thought blocker

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Student12 said:
Question 27 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

Question 29
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
Click to expand...
27)
find d which is 1/n -->1/500 * 10^ (-3) = 2 x 10^-6.
then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D

29)
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
 
Princess Raven

Princess Raven

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Thought blocker said:
s12_s12
17)
At the bottom height is zero so mgh = zero
now going upwards..
2 block has height = 1 so mg1h
3 block = mg2h
4 block = mg3h
Sum up P.E ans is 6mgh

29)
c = f * λ
We know f = 1 / t
so c = 1 / t * λ
t = λ / c
now it took 3 wavefronts to reach XY to P
so t = 3λ / c

30)
Its a standing wave, so the 33cm is the distance between two nodes. The wavelength thus becomes .66m. F = 330/0.66 = 500. T = 1/500 = 0.002s = 2ms. B is the right answer

s07_1
Suchal Riaz
me gtg :)
Click to expand...
Thank u :) just one question y do u do distance between 2 nodes I mean isnt the normal way just 33 cm?
 
Princess Raven

Princess Raven

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Suchal Riaz said:
3:
simple. the speed must have units m/s
in (a) √(gλ)=(m²/s²)^½ = m/s
no need to check rest of them :p
7:
O level vectors: x=v-u
v-u = at
so x=at

20:
All i can do is visualise which parts are bending in what way. X and Y are being stretched but Z is being pressed/compressed.
X and Y experience tension while Z experiences compression. just visualise.
40:
the particle which completes acceleration with the lowest speed must have had lowest acceleration as the distance is same.
F is not the same. force directly proportional to charge. acceleration = F/m so acceleration will depends on charge and mass. A ∝charge/mass
so find charge to mass ratio. the one with smallest ratio has lowest acceleration therefore lowest speed at the end.
(a)1/1=1
(b)2/4=0.5
(c)3/7=0.428...
(d)4/9=0.44...
smallest is (c)
Click to expand...
OMG THANK U!! ur explanations rock! Seriously I get everything now.. So for the last one q40 we always do charge/mass ratio? For the lowest speed questions I mean?
 
K

Kamihus

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Thought blocker said:
32)
The arrows point downwards which means the charge at the top is positive and the bottom is negative. So, the +Q will be attracted downwards and the -Q will be attracted upwards, spinning the rod anti-clockwise. The resultant force is zero because F = E/Q and the Q (both charges) are equal but opposite. So resultant force is zero. If the charges weren't the same in magnitude, then we would have a resultant force.

14)
sagar had explained v.well..
Suppose we take the upward direction to be positive, and the downward direction to be negative, we can again write the momentum equations and use them to find the final velocity of the system - the forces between the clay and the lead pellet are huge compared to other external forces, so even though there are external forces acting on the system during the collision, we can assume the momentum stays approximately constant.

So, the initial momentum is the momentum of the bullet alone, which is equal to mv = (5.0/1000) * 200 = 1.0 kg ms^-1 . The clay block is stationary, so it does not contribute any momentum to the system initially.
When the bullet collides with the clay, the bullet gets stuck in the clay and they both move off with the same velocity, which we'll call v(f). The mass of the lead pellet+ the clay block = (95/1000 + 5/1000) = (100/1000) = 0.1 kg.
Their final velocity = v(f)
Therefore, the final momentum of the system is 0.1 * v(f)

Since this is equal to the initial momentum, we can write (0.1 kg) * v(f) = (1.0 kg ms^-1) and so v(f) = 10 ms^-1.

This is the speed with which they rise after the collision. While the system moves upwards, gravity accelerates them downwards at a constant rate of -9.81 ms^-2. When they reach the highest point above the original position, their velocity is 0. Therefore, we can use the formula v^2 = u^2 + 2as to give us "s", which is the maximum height. So,

(0)^2 = (10)^2 +2(-9.81)s
19.62 * s = 100
s = 5.09 m = 5.1 meters = A.
Click to expand...
(y)
You should add some index in your thread of all the difficult questions that you've answered as they are asked again and again.
 
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not.maria said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
kindly help me out in
q 10 21 26 38
Click to expand...
10)
At X, sand is stationary and truck is moving so when they stick together the speed decreases.
At Y, sand also leaves with some speed and as the total mass remains same, the speed will not change.

21)
--> extension of P is half that of Q, as you can see for more stress there is less strain for P that Q, hence statement 1 is incorrect.
--> young modulus=stress/strain for P graph is steeper than Q, means greater young modulus
--> for statement 3, before 0.1, it is a straight line

26)
As shown in the fig, at time = 18 s, phase diff btw the 2 waves = 180° or π.
1 wavelength = 360°
So, 1/8 of a wavelength = 360°/8 → 45°.

Phase diff = 180° at 18 s.
So, phase diff = 45° at x.
x = (18 * 45)/180
x = 4.5 s

Ans = B


38)
find total resistance of voltmeter and resistor connected in parallel by 2oo x 200/400=100ohms
then find the share of this resistance in total batterys voltage by :
100/500(total resistance) x 60=12V,since both voltmeter and resistor is in parallel each is having 12 V.
 
A

amal sharkawi

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A projectile is launched at 45° to the horizontal with initial kinetic energy E.

Assuming air resistance to be negligible, what will be the kinetic energy of the projectile when it
reaches its highest point?
A 0.50 E B 0.71 E C 0.87 E D E
can any one answer this plz??
 
Thought blocker

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amal sharkawi said:
A projectile is launched at 45° to the horizontal with initial kinetic energy E.

Assuming air resistance to be negligible, what will be the kinetic energy of the projectile when it
reaches its highest point?
A 0.50 E B 0.71 E C 0.87 E D E
can any one answer this plz??
Click to expand...
vertical velocity= u cos 45
horizontal velocity = u sin 45
so, initial K.E. : the vertical and horizontal components of kinetic energy = 1/2 m ( u cos 45 ) + 1/2 m ( u sin 45 ) ......... ( 1/2 m v^2 formula )

so, sin 45 = 0.71
....cos 45 = 0.71 ( check them in the calculator )

so they will give the same kinetic energy for both velocities ( horizontally and vertically )

so at the highest point, (theta) = 90 ( for the vertical velocity ) , horizontal velocity doesn't change
so final kinetic energy = 1/2 m ( u cos 90 ) + 1/2 m ( u sin 45 )..................... ( horizontal velocity is constant ) ( cos 90 is 0 )
final kinetic energy = 1/2 m ( u sin 45 ) + 0 only which is half of the initial kinetic energy.
 
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