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Ans is D, Because:

^R/R = ^I/I + ^V/V----- ^= Delta

R=V/I

R= 15/2=7.5

SO,

^R/R=^I/I+^V/V

^R=(^I/I+^V/V)R

^R=(0.2/2 + 0.5/15)*7.5 = 2

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Ans is D, Because:

^R/R = ^I/I + ^V/V----- ^= Delta

R=V/I

R= 15/2=7.5

SO,

^R/R=^I/I+^V/V

^R=(^I/I+^V/V)R

^R=(0.2/2 + 0.5/15)*7.5 = 2

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^R/R = ^I/I + ^V/V----- ^= Delta

R=V/I

R= 15/2=7.5

SO,

^R/R=^I/I+^V/V

^R=(^I/I+^V/V)R

^R=(0.2/2 + 0.5/15)*7.5 = 2

It's D.

The uncertainty in R can find by first finding the fractional uncertainties of I and V, and then adding them and multiplying by the the value of resistance calculated.

( 0.2/2 + 0.5/15 ) * 7.5

= -+1 ohm

thank you!

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In this question we're told that the velocity is positive upwards. Also, we know that at highest & lowest points the mass will come momentarily to rest. So when mass is at lowest point, the velocity will be zero, and afterwards, when it'll again start moving upwards, it's velocity will be positive. this elimintes the chance of B being the answer as after coming to rest, the velocity is negative showing that the ball is moving downwards now.View attachment 58723

can someone tell me why the ans isnt B???! :'(

Hope you understood.

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So is C the ans?In this question we're told that the velocity is positive upwards. Also, we know that at highest & lowest points the mass will come momentarily to rest. So when mass is at lowest point, the velocity will be zero, and afterwards, when it'll again start moving upwards, it's velocity will be positive. this elimintes the chance of B being the answer as after coming to rest, the velocity is negative showing that the ball is moving downwards now.

Hope you understood.

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No. As I explained above, it'll be mometarily at rest at lowest point, so it's velocity will be zero there. So it can't be C.So is C the ans?

It's D.

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In this question we're told that the velocity is positive upwards. Also, we know that at highest & lowest points the mass will come momentarily to rest. So when mass is at lowest point, the velocity will be zero, and afterwards, when it'll again start moving upwards, it's velocity will be positive. this elimintes the chance of B being the answer as after coming to rest, the velocity is negative showing that the ball is moving downwards now.

Hope you understood.

yup! got it

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Let m√2gh1 = V1 and -m√2gh2= V2View attachment 58724

and why is this B not A?? when it changes direction isnt it supposed to be subtracted?

So change in momentum will be,

V1-V2

which is equal to

m√2gh1-(-m√2gh2)

so,

m√2gh1+m√2gh2

is the ans.

Hope u got that

Last edited:

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Sorry for getting late.View attachment 58724

and why is this B not A?? when it changes direction isnt it supposed to be subtracted?

Look, if you take the downward direction to be negative, and the upward direction to be positive, then the momentum before collision will be : -m√(2gh1)

and the collision after collision will be m√(2gh2)

so then the change in momentum can be found by:

final momentum - initial momentum

= m√(2gh2) - (-m√(2gh1))

= m√(2gh2) + m√(2gh1))

So Answer is B.

P.S. You can also take the upward direction to be -ive and downward direction to be +ive, but this will result in the change in momentum to be :

- m√(2gh2) - m√(2gh1)) <--- this will be also correct, but this not an option given here.

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Let m√2gh1 = V1 and -m√2gh2= V2

So change in momentum will be,

V1-V2

which is equal to

m√2gh1-(-m√2gh2)

so,

m√2gh1+m√2gh2

is the ans.

Hope u got that

Sorry for getting late.

Look, if you take the downward direction to be negative, and the upward direction to be positive, then the momentum before collision will be : -m√(2gh1)

and the collision after collision will be m√(2gh2)

so then the change in momentum can be found by:

final momentum - initial momentum

= m√(2gh2) - (-m√(2gh1))

= m√(2gh2) + m√(2gh1))

So Answer is B.

P.S. You can also take the upward direction to be -ive and downward direction to be +ive, but this will result in the change in momentum to be :

- m√(2gh2) - m√(2gh1)) <--- this will be also correct, but this not an option given here.

oh yeah...i got it!! thanx a lot

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.

The weight W of the climber is 520 N. The rope, of negligible weight, is attached to the climber and to a fi xed point P where it makes an angle of 18° to the vertical. The reaction force R acts at right angles to the wall. The climber is in equilibrium. (a) Complete Fig. 2.2 by drawing a labelled vector triangle to represent the forces acting on the climber.

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Anyways, this question was solved by myself.

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Please guide me some good A Levels teachers in Karachi

Thanks in advance

Thanks in advance

Actually I'm in a really confused state, my problem is related to M1 mechanics and physics....Alright, so we all know that when a simple object like a block of wood moves in a forward direction there is kinetic friction between the block and the floor in the opposite direction.... But things get tricky when it comes to an individual wheel, my mind is actually boggled about *

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u = 1.66 x 10^-27http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_43.pdf question 1bii) HELP PLEASE! I don't understand how they got mass as 1.66x10^-7 (other than that I understand everything) thanks a lot in advance!

Mass of hellium is 4u = 4 x 1.66 x 10^-27

Like 100km = 100000m.

thank you so, so much!!u = 1.66 x 10^-27

Mass of hellium is 4u = 4 x 1.66 x 10^-27

Like 100km = 100000m.