Physics: Post your doubts here!

[Thought blocker]

thnks

 

[Hadi Murtaza]

?

 

[Thought blocker]

?

How was your Paper 4 ?

 

[Hadi Murtaza]

How was your Paper 4 ?

Im doing A2 Edecxel

 

[Hadi Murtaza]

Im doing A2 Edecxel

They have units, not pprs

 

[Thought blocker]

Oh, okay.

 

[Zepudee]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf

Q8,9,13,18,26 Please help me

 

[Champ101]

Determine the gradient of the line of best fit. Include the uncertainty in your answer

for this question: we just subtract the gradient of worst fit- gradient of best fit to get the value of uncertainty?

 

[Student12]

How do you calculate the absolute uncertainty for Xc ? This is May/June 2010_ 51

 

[Student12]

The absolute uncertainty for this?

 

[sagar65265]

Determine the gradient of the line of best fit. Include the uncertainty in your answer

for this question: we just subtract the gradient of worst fit- gradient of best fit to get the value of uncertainty?

Yeah, that should be it. There is an alternate method, though, just in case you're interested:

Find the gradients of the steepest line that passes through all the error bars, and the shallowest line that passes through all the error bars.

Suppose they happen to be 5.7 and 4.3 in any units (and the gradient of the best fit is 4.0).

The calculation you carry out = 0.5 * (Gradient of Steepest line - Gradient of shallowest line)

In our example, this is 0.5 * (1.4) = 0.7 (in the relevant units).
So we'll write that the gradient is 4.0 ± 0.7.

The marking scheme from a 2012 October/November paper probably puts this best:

"Uncertainties in Question 2

(c) (iii) Gradient [U2]

Uncertainty = gradient of line of best fit – gradient of worst acceptable line

Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)"

So really, either method is accepted as long as you make it clear which one you are using.

Good Luck with all your exams!

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf

Q8,9,13,18,26 Please help me

Q.8
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(2m)(u) + (m)(-u) = (2m)(v₁) + (m)(v₂)
2mu - mu = 2mv₁ + mv₂
mu = 2mv₁ + mv₂
u = 2v₁ + v₂ . . . . . (i)

e = 1 , as collision is elastic
e = (v₂ - v₁)/(u₁ - u₂)
1 = (v₂ - v₁)/(u₁ - u₂)
v₂ - v₁ = u₁ - u₂
v₂ - v₁ = u - (-u)
v₂ - v₁ = 2u
v₂ = 2u + v₁

Substitute in (i)
u = 2v₁ + (2u + v₁)
u = 3v₁ + 2u
3v₁ = -u
[ v₁ = -u/3 ]

v₂ = 2u + v₁
v₂ = 2u - u/3
[ v₂ = 5u/3 ]

So answer: A

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf

Q8,9,13,18,26 Please help me

Q.9
Impulse (or change in momentum) = Force * time
Impulse = 60 * 0.5 = 30 Ns

Initial momentum of trolley = 30 kg * 3 m/s= 90 Ns
Impule (or change in mometum) = 30 Ns
So final momentum = 90 - 30 = 60 Ns
Final speed = final momentum/mass
Final speed = 60/30 = 2 m/s

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf

Q8,9,13,18,26 Please help me

Q.18
Pressure = F/A = W/A = mg/A = g * m/A
m/A = Pressure/g

Density = m/V = m/A * 1/h
Density * h = m/A
Denisty * h = Pressure/g
Denisty = (Pressure/g) /h
Density = Pressure/(g * h)
Density = 15 200/(10 * .8)
Density = 15 200/8
Density = 1900 kg/m³
Answer: B

 

[Thought blocker]

Q13 ?

 

[Hadi Murtaza]

Q8,9,13,18,26 Please help me[/quote]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf

Q8,9,13,18,26 Please help me

Q.26
When charge Q is at position of q. It exerts a force of da electric field strength on da charge n its magnitude depends upon da strength of da electric field n da amount of charge on Q.
F = Eq
E = F/q

 

[Thought blocker]

Q8,9,13,18,26 Please help me

Q.26
When charge Q is at position of q. It exerts a force of da electric field strength on da charge n its magnitute depends upon da strength of da electric field n da amount of charge on Q.
F = Eq
E = F/q[/quote]
I thought it was answered by Zep.. something Copycat

 

[Hadi Murtaza]

Q13 ?

Im kindof lousy at explaining dis 1

Initial E = E[x] + E[y]
Final E = E[x] only
So Final E = (1/2) Initial E

 

[Thought blocker]

Im kindof lousy at explaining dis 1

Initial E = E(x) + E
Final E = E(x) only
So Final E = (1/2) Initial E

Projectile me aisa sab ata hain ?

 

[Hadi Murtaza]

Q.26
When charge Q is at position of q. It exerts a force of da electric field strength on da charge n its magnitute depends upon da strength of da electric field n da amount of charge on Q.
F = Eq
E = F/q

I thought it was answered by Zep.. something Copycat [/quote]
Wat