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Physics: Post your doubts here!

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24) the youngs modulus, E is the same for the same material, regardless of the length and cross sectional area, etc.
the SAME STEEL is used, so E is the same

26) ok now the molecules at Q are 2r from the source
and Intensity is inversely proportional to the distance from the source
d = 2r >> Intensity x 1/ 2^2 = 1/4
if intensity is decreasing by 4, and intensity is directly proportional to the amplitude...
then the amplitude is decreased by the square root of 4 = 2, so amplitude x 1/2
amplitude Q = 8 x 1/2 = 4 micrometers

please answer this: n10 12: qs 8(my answer is coming 0.49ms-2) can any1 help me plzz
 
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please answer this: n10 12: qs 8(my answer is coming 0.49ms-2) can any1 help me plzz

ok when at Y, the body has traveled 40 m, in 12 seconds so the speed at Y is v = d / t = 40 / 12 = 3.3 ms^-1
this is going to be the initial speed for the length YZ
the speed at Z = d / t = 40 / 6 = 6.6 ms^-1
this is the final speed for the length YZ
now use v^2 = u^2 + 2as for YZ
(6.6)^2 = (3.3)^2 + 2 a (40)
a = 0.36 ms^-2
 
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for 18 : E= power output / input ... out put is already given ...for input you multiply fuel comsumption by energy content ..convert to SI units. to check wether you did it correct multiply with units and you'll get ______j /s that is the unit of power . (sorry i think this is not what you asked for , but anyhow might be useful for sm1 else)


question 24: since both wires of same material so E young modulus is same
equate equations for both
F is same
so , FL/ pi r^2 x1 = F 3L/ 4 pi r^2 x2 ........................you will get x2= 3/4 x1

question 34 :
 

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for 18 : E= power output / input ... out put is already given ...for input you multiply fuel comsumption by energy content ..convert to SI units. to check wether you did it correct multiply with units and you'll get ______j /s that is the unit of power . (sorry i think this is not what you asked for , but anyhow might be useful for sm1 else)


question 24: since both wires of same material so E young modulus is same
equate equations for both
F is same
so , FL/ pi r^2 x1 = F 3L/ 4 pi r^2 x2 ........................you will get x2= 3/4 x1

question 34 :
thanks...:D
 
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ok when at Y, the body has traveled 40 m, in 12 seconds so the speed at Y is v = d / t = 40 / 12 = 3.3 ms^-1
this is going to be the initial speed for the length YZ
the speed at Z = d / t = 40 / 6 = 6.6 ms^-1
this is the final speed for the length YZ
now use v^2 = u^2 + 2as for YZ
(6.6)^2 = (3.3)^2 + 2 a (40)
a = 0.36 ms^-2
you get 0.4 ms not 0.36 ?!
 
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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
question 10 need help urgently~!
From my knowledge, momentum is constant if there's no external force acting on a system, so p = mv for the truck, since p is constant, when m increases v decreases, but why does the velocity stays the same when the sand is released from the truck? (Mass of the truck decreases?)


first of all newtons first law applies here .....only when a force acts on the object it causes a change in its velocity
so when sand is added first v decreases due to increase in mass ......when sand falls out the velocity stays the same since there is no force to increase its velocity later ...it tend to continue its motion along the same path with same speed
 
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first of all newtons first law applies here .....only when a force acts on the object it causes a change in its velocity
so when sand is added first v decreases due to increase in mass ......when sand falls out the velocity stays the same since there is no force to increase its velocity later ...it tend to continue its motion along the same path with same speed
I've got it from my friend, i think his explanation sounds more reasonable :
when mass is added, velocity decreases due to the increase in mass (p = mv), where p is a constant,
when the sand is realeased, looking a the truck itself there's loss in mass, but if we look at the sand and the truck as a system,
p = (m1 + m2)v (m1 = mass of sand, m2 = mass of truck)
so when the sand is released, the sand continues to move with velocity v, so thus the truck, as we can see from
p = m1v + m2v (expanding the equation)
So momentum is still conserved throughout the system!
Hope this helps!
 
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Why does the height increase again ? Wont it be the same height. I agree the height increases by "h" in the right tube. but after that i dont understand.
U got trolled by the one who set this question lol, (me too actually xD)
u see, when the right tube raise by a height h, the left tube too will raise through a height h, (imagine the water flowing from the left tube to the right tube)
so the change in height is acutally 2h
 
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Why does the height increase again ? Wont it be the same height. I agree the height increases by "h" in the right tube. but after that i dont understand.
the liquid has to move. If liquid increase in one tube, it has to come from somewhere which is the another tube. So as liquid increase, liquid decrease kinda thing....Another example, if you have 5 apples each in two basket and you had to move 2 apples from one basket to another, one basket has 7 and another will only have 3 left. 7-3=4 .....net difference is 4
 
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Oct09
Q12. since the btm belt is slacked, so tension is only at the upper belt, torque at Q = 3.0 N m,
so 3.0 = T x (100 x 10^-3)/2
T = 60N
Torque at P = T x (150 x 10^-3)/2 = 4.5
Q13. assuming it is projected with velocity v, so the initial KE, E= (mv^2)/2
at the highest point, the vertical component of the veolcity = 0, so its left with v cos 45,
so final KE = 1/2 x m x (v cos 45)^2 = 0.5E
Q14. By conservation of momentum, 2 x 2 = 1 x v, v =4
so the 1kg trolley moves 4ms-1 to the right,
calculating the total KE of the both trolleys, which is 1/2 x 2 x 2^2 + 1/2 x 1 x 4^2 = 12J
assuming no energy is lost, the energy stored in the spring is converted to KE gained by the trolleys, so its 12J

Oct11/12
Q23. the area under graph is the elastic PE, so we can c that the answer is obviously (i hope its obvious for u too) B. not much that i could explain here
 
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the liquid has to move. If liquid increase in one tube, it has to come from somewhere which is the another tube. So as liquid increase, liquid decrease kinda thing....Another example, if you have 5 apples each in two basket and you had to move 2 apples from one basket to another, one basket has 7 and another will only have 3 left. 7-3=4 .....net difference is 4

Then left tube will have height of x-h. and the right tube will have x + h right???? and now how come the difference is 2h ?
 
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Oct09
Q12. since the btm belt is slacked, so tension is only at the upper belt, torque at Q = 3.0 N m,
so 3.0 = T x (100 x 10^-3)/2
T = 60N
Torque at P = T x (150 x 10^-3)/2 = 4.5
Q13. assuming it is projected with velocity v, so the initial KE, E= (mv^2)/2
at the highest point, the vertical component of the veolcity = 0, so its left with v cos 45,
so final KE = 1/2 x m x (v cos 45)^2 = 0.5E
Q14. By conservation of momentum, 2 x 2 = 1 x v, v =4
so the 1kg trolley moves 4ms-1 to the right,
calculating the total KE of the both trolleys, which is 1/2 x 2 x 2^2 + 1/2 x 1 x 4^2 = 12J
assuming no energy is lost, the energy stored in the spring is converted to KE gained by the trolleys, so its 12J

Oct11/12
Q23. the area under graph is the elastic PE, so we can c that the answer is obviously (i hope its obvious for u too) B. not much that i could explain here
Thank you so much! You are a saviour for dumb people like me !! :D
 
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U got trolled by the one who set this question lol, (me too actually xD)
u see, when the right tube raise by a height h, the left tube too will raise through a height h, (imagine the water flowing from the left tube to the right tube)
so the change in height is acutally 2h

Why would the left tube will also raise by h ? wont it decrease by h ?
 
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