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Physics: Post your doubts here!

Rockstar RK

Rockstar RK

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Ahmed Aqdam said:
Charge=Force/Electric field strength
Mass=Force/acceleration
Force will cancel so E/g

Charge is negative as weight acts downwards and as it is in equilibrium an upward force acts to counterbalance it. It will be towards positive charge so drop is negative.
Click to expand...
Please anwer this:
A moving body undergoes uniform acceleration while travelling in a straight line between points
X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and
from Y to Z is 6.0 s.
What is the acceleration of the body?
A 0.37 m s–2 B 0.49 m s–2 C 0.56 m s–2 D 1.1 m s–2
 
not.maria

not.maria

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Thought blocker said:
11)
The angle that T1 makes with R is greater than the angle that T2 makes with R.
R has to be balanced by the horizontal componts of T1 and T2
Let the angle be θ
R= T1cosθ + T1cosθ and R= T2cosθ + T2cosθ
We can see that for cos,, the greater the angle.. less the value of cosθ
so in order to cancel out R the value of T1 must be greater as cos θ is less in diagram 1. Therefore T1>T2

28)
F= Eq
F = 3 x 10^7 x 1.6 x 10 ^(-19)
F = A

the q is always the charge of the charge that is moving, not the charge of the electric field

39)
It can not be A cause there is suppose to be a deflection not B cause they weight the same not D cause alpha particles are postive and so is the gold necule they repel and not attract so it is JUST C
Click to expand...
just to clear things up:)
the ans to 39 is B not C
you have written C and explained for C to be correct
Markscheme
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_ms_12.pdf
 
ZaqZainab

ZaqZainab

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not.maria said:
why not A or C
Click to expand...
A has no deflection for the second one and that is wrong as both the gold nucleus and alpha particle carry + change there is suppose to deflection
C as i said the closer you get to the nucleus the stronger the push
in C both the particles are deflected with the same angle which is wrong as the one on the top is further away from the nucleus while the one on the bottom is closer
 
Thought blocker

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crazytaylorfanXD said:
i need help in a few questions from
NOV 12 P11 Qs ( 6 , 11 , 23 , 29 , 36) :/
Click to expand...
6)
The question states that "X is measured with a percentage uncertainty of ±1 % of its value at all temperatures." meaning that the percentage uncertainty is unchanged, so A and B is eliminated.
then you need to apply some logic here, will the actual uncertainty get bigger as you approach 100 or 0?.
you can quickly do some math here, 100x1% = ±1 uncertainty while 0x1% = 0
so answer is C, least actual uncertainty when temperature close to 0

11)
The object hits the wall with a speed v, and rebounds with the same speed, but a velocity in the opposite direction.
Kinetic Energy does not depend on the direction of an object's velocity, just the magnitude of that objects velocity, i.e. the speed. Therefore, since the speed remains the same even after the collision, the Kinetic Energy of the object is conserved.

By the same logic applied above, we can say that the speed is conserved, so that option is eliminated.

Lastly, the mass of the object doesn't change, and therefore the mass also remains conserved.
By elimination, we can confidently say that the momentum of the object is the only value that is not conserved - momentum, unlike the other values, is a vector quantity; since the direction of the object's speed changes, the momentum also changes, and is therefore not conserved.

23)
We use F = kx
so we know k for P and Q is k and for R is 3k
--> Therefore total extension of P and Q = k + k = 2k and extension for P,Q and R = 1/2k + 1/3k = (5/6)^-1k = 6/5k
so F = W, k = 6/5k and x = ?
W = 6/5k * x
x = 5w/6k

29)
There should be more points on to which the light strikes so it's either B or C, but hence in C the lines are closer, there is more intensity

36)
Firstly the ammeter is directly connected with the fixed resistor so there is no change in readings.
And its obvious, when there are two resistors, if one resistor's resistance is increasing, so others resistor's resistance would decrease so R is proportional to V hence V decreases.
 
Thought blocker

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Rockstar RK said:
A moving body undergoes uniform acceleration while travelling in a straight line between points
X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and
from Y to Z is 6.0 s.
What is the acceleration of the body?
A 0.37 m s–2 B 0.49 m s–2 C 0.56 m s–2 D 1.1 m s–2
Click to expand...
Dushyant, here I took distance XY and XZ and solve them simultaneously.
This was the only way I could think of :p
ZaqZainab I tried your way now :p LOL
Anyways here is my method :¬

Use the equation of motion d = ut + 0.5at^2

equation 1 : 40 = 12u + 72a

equation 2 : 80 = 18u + 162a

[I added the distances, so that initial speed of both equations will be same]

now solve these simultaneously, and you will get 0.37 m/s^2 as the answer.

P.S. I solved it in full detail, go some few pages back, and you'll get your answer :)
 
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forever_chocoholic said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
Q:22
please explain! :(
Click to expand...
Case i)
When all 3 springs are attached to the load W :
So, think.. If you have 3 springs and you have 1 weight attach to them what will be the force for each string ?
It would be W/3
So we know F = kx
k = W/3x

Case ii)
When middle spring is removed and 2W load is attached to them :
So, think again, what would be the force for each string ?
It would be W
So you know the value of k i.e is W/3x and constant doesn't change, as it is named as constant itself.
so considering :
F = kx
x = W / (W/3x)
x = ( W * 3x ) / W
so W gets canceled off and now you are remained with 3x which is our answer, D.
 
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