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Physics: Post your doubts here!

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MJ 2010 42 Q1

2R above earth's surface ---->3R from the centre of the earth.
So, potential at 3R from center of the earth = 2.1 * 10^7 J/kg

v knw tht change in kinetic energy = change in potential energy
so, 1/2 * m * v^2 = mgh ----> φ * m
1/2 v^2 = φ
v^2 = (2.1 * 10^7) * 2
v = 6.5 * 10^3 m/s

Hope u gt it!
 
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2R above earth's surface ---->3R from the centre of the earth.
So, potential at 3R from center of the earth = 2.1 * 10^7 J/kg

v knw tht change in kinetic energy = change in potential energy
so, 1/2 * m * v^2 = mgh ----> φ * m
1/2 v^2 = φ
v^2 = (2.1 * 10^7) * 2
v = 6.5 * 10^3 m/s

Hope u gt it!

oh !! just got confused with the 2R and 3R thing..Tricky CIE...silly ME :p
 
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Can someone please help me with question 2a (i) and (ii)? I can never understand the difference between potential and potential energy.
Thank you.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_4.pdf

2 a i) Gravitational potential energy = gravitational potential * mass ---> gravitational potential = GM/R.
So, Gravitational potential energy = GM1M2/R
M1 = mass of the gold nucleus = (197 * 1.66 * 10^-27) kg
M2 = mass of the α nucleus = (4 * 1.66 * 10^-27) kg
energy = {(6.67 x 10^-11) x 197 x 4 x (1.66 x 10^-27)^2}/9(.6 x 10^-15)
energy = 1.51 x 10^-47 J
 
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ii) Similarly, electric potential energy = kQ1Q2/R
Q1 = charge of gold nucleus = 79 * 1.6 * 10^-19 C
Q2 = charge on α nucleus = 2 * 1.6 * 10^-19 C

energy = [79 * 2 * (1.6 * 10^-19)^2] / (4 * π * 8.85 * 10^-12 * 9.6 *10^-15)
energy = 3.79 * 10^-12 J

Hope u gt it!!
:D
 
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ii) Similarly, electric potential energy = kQ1Q2/R
Q1 = charge of gold nucleus = 79 * 1.6 * 10^-19 C
Q2 = charge on α nucleus = 2 * 1.6 * 10^-19 C

energy = [79 * 2 * (1.6 * 10^-19)^2] / (4 * π * 8.85 * 10^-12 * 9.6 *10^-15)
energy = 3.79 * 10^-12 J

Hope u gt it!!
:D
Yes, I do get the electrical potential energy part.
Although, I'm not sure when to use both masses as in GMm/r or to just use one mass as in GM/r. Sometimes, when planets are involved they just want you to use one mass.
But thank you so much. :) It helped a lot and Jazak Allahu Khairan.
 
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2 a i) Gravitational potential energy = gravitational potential * mass ---> gravitational potential = GM/R.
So, Gravitational potential energy = GM1M2/R
M1 = mass of the gold nucleus = (197 * 1.66 * 10^-27) kg
M2 = mass of the α nucleus = (4 * 1.66 * 10^-27) kg
energy = {(6.67 x 10^-11) x 197 x 4 x (1.66 x 10^-27)^2}/9(.6 x 10^-15)
energy = 1.51 x 10^-47 J
Here, you calculated energy as 1.51 x 10^-47 Joules but isn't that the gravitational potential, not energy. Because then to get energy you would need to multiply this energy by a mass. Right? And what's the difference between electric potential and electrical potential energy?
 

Moi

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Can someone please tell me how to find the time in Summer 11/paper 2/ variant 3/ Q- 3 -Di !! The ms only gives the answer not the way ! am completely stuck !
Also, in the same paper, how to do Q6-b ?? how to solve it and know where to start the sketch of the wave ?

Thank you, please answer as soon as possible !
 
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guys how do no how many significant figures an answer can hav. I lose all my marks cus of this!:(
 

Moi

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guys how do no how many significant figures an answer can hav. I lose all my marks cus of this!:(

usually it's 3 sig. figures. unless the answer had more, then you'll write the whole thing or use a standard form (ex: 4.31x10^6 )
 

Moi

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Can someone please tell me how to find the time in Summer 11/paper 2/ variant 3/ Q- 3 -Di !! The ms only gives the answer not the way ! am completely stuck !
Also, in the same paper, how to do Q6-b ?? how to solve it and know where to start the sketch of the wave ?

Thank you, please answer as soon as possible !

Can someone please answer :)
 
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In Youngs double slit experiment, decreasing the light intensity on on of t he slits decreases the contrast b/w light and dark fringes. Does increasing the light intensity have the same effect?
 
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Help please. I am self taught and I don't understand how to label diagrams. View attachment 25174View attachment 25175

For part i draw a line joining another set of points of intersection (preferably the one right next to line X so that its easier for me to explain part ii)
For part ii, draw a line exactly halfway b/w line X and the line for part i... Since destructive interference occurs in b/w two constructive interference regions.
 
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