Physics: Post your doubts here!

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf

Q.5
X = V*cos(Θ)
Y = V*sin(Θ)

Maximum value of cos(Θ) is when Θ = 0 and Minimum when Θ = 90° , so as Θ increases from 0° to 90°, V*cos(Θ) decreases, therefore X decreases.
Maximum value of sin(Θ) is when Θ = 90 and Minimum when Θ = 0° , so as Θ increases from 0 to 90°, V*sin(Θ) increases, therefore Y increases.

 

[Hadi Murtaza]

Q.5
X = V*cos(Θ)
Y = V*sin(Θ)

Maximum value of cos(Θ) is when Θ = 0 and Minimum when Θ = 90° , so as Θ increases from 0° to 90°, V*cos(Θ) decreases, therefore X decreases.
Maximum value of sin(Θ) is when Θ = 90 and Minimum when Θ = 0° , so as Θ increases from 0 to 90°, V*sin(Θ) increases, therefore Y increases.

Therefore answer is C

 

[DeViL gURl B)]

Therefore answer is C

Thank u

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf

Help in question 7 and 27 pleaseeee !!
Thank u

 

[Hadi Murtaza]

Thank u

Ur mooossst welkum
U understood ? any doubts ?

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf

Help in question 7 and 27 pleaseeee !!
Thank u

Hadi Murtaza help would be well appreciated

 

[Hadi Murtaza]

Hadi Murtaza help would be well appreciated

Sure gimme 15 mins, abit busy now

 

[DeViL gURl B)]

Sure gimme 15 mins, abit busy now

Sure thanks

 

[Hadi Murtaza]

Sure thanks

Q.7
Micrometer scale is divded into 0.5 mm readings, n der r 5 scale lines visible, so diameter of wire = 0.5 * 5 = 2.5 mm.
nearest answer comes to B, 2.45 mm.

 

[M.Shahzaib Shoaib]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

Question 16 & 17 !

Complete explanation needed ..

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

Question 16 & 17 !

Complete explanation needed ..

Q.16
When a ball is thrown up in da air, its P.E. increases and its K.E. decreases until it reaches maximum height, however da sum of K.E. n P.E. remains constant throught da flight. Answer: B

Q.17
V(x) = v V = 2v
M(x) = 2m M = 2

K.E of x = (1/2) * 2m * (v)² = mv²
K..E of y = (1/2) * m * (2v)² = (m/2) * 4v² = 2mv²

Therefore K.E. of y = 2 * K.E. of x
or K.E. of x = (1/2) * K.E. of y

Hence answer is : A

 

[M.Shahzaib Shoaib]

Q.16
When a ball is thrown up in da air, its P.E. increases and its K.E. decreases until it reaches maximum height, however da sum of K.E. n P.E. remains constant throught da flight. Answer: B

Q.17
V(x) = v V = 2v
M(x) = 2m M = 2

K.E of x = (1/2) * 2m * (v)² = mv²
K..E of y = (1/2) * m * (2v)² = (m/2) * 4v² = 2mv²

Therefore K.E. of y = 2 * K.E. of x
or K.E. of x = (1/2) * K.E. of y

Hence answer is : A

Thanks

 

[Hadi Murtaza]

Thanks

U r most welkum

 

[Student12]

Paper 5 people.. all prepared ? :\

 

[Snackbox86]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
Can i have help with q15 2011w p12
Thank you

 

[Champ101]

Paper 5 people.. all prepared ? :\

Ahh no idea!! just solving some papers! you??

 

[Champ101]

Uncertainty question:

If R= 4.00±0.05, find R^2 including absolute uncertainty? i am not sure can anyone do it?

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
Can i have help with q15 2011w p12
Thank you

Energy = Force * distance
E = Fs

When force 'F' is applied for a distance 's' , KE increases by 4 J or + 4 J
so
KE = F * s
Fs = 4 J . . . . . . (i)

When force '2F' is applied for a distance '2s', KE increases by 'x' , where x is unknown.
KE = 2F * 2s
4Fs = x
Substitute Fs from (i)
x = 4(4) J
[ x = 16 J ]

So KE increases by 16 J n becomes, 4 + 16 = 20 J

Answer: B

 

[Champ101]

When R is raised by a power 'n' , da uncertainty is multiplied by 'n'

So
R² = (4.00)² ± 2*(0.05)
R² = 16.00 ± 0.1

yup i got it too, but the ms says its From ± 0.4 to ± 0.6 ?