Physics: Post your doubts here!

[ZohaibAsad]

I am having some problems answering summer 2008 paper 2 question (4c) . How do I get the the error?

Post the link plz

 

[InnocentAngel]

Energy of photon = work function of a metal + Maximum kinetic energy of the electron
E = Φ + K.E
hc/λ = Φ + K.E

Where c is the speed of light ( 3 x 10^8 m/s)

---------------------------------------------------
A surface is eradiated with lights of wavelength 5.5 x 10^-7m and electrons are just able to escape from the surface.

We can deduce that, just able to escape means K.E = 0
hc/λ = Φ + K.E
hc/5.5 x 10^-7 = Φ + 0 ------- (1)

When light of wavelength 5.0 x 10^-7m is used , electrons emerge with energies of up to 3.6 x 10^-20 J.
hc/λ = Φ + K.E
hc/5.0 x 10^-7 = Φ + 3.6 x 10^-20 ---------- (2)

(1) - (2)
hc/5.5 x 10^-7 - hc/5.0 x 10^-7 = - 3.6 x 10^-20
.
.
.
Solving... h = 6.6 x 10-34

now im understng! so its the k.e which is the 3.6 x 10^-20 . I thought it was the total of the work function and the k.e. Thanks lot!

 

[Huzza]

can u help me with this question:
a skateboard of mass 4kg is moving at asteady speed of 2m/s. Calculate its momentum.
A 1kg bag of sugr is then dropped vertically onto it. Calculate its new speed assuming momentum is unchanged.

 

[memorypriest]

can u help me with this question:
a skateboard of mass 4kg is moving at asteady speed of 2m/s. Calculate its momentum.
A 1kg bag of sugr is then dropped vertically onto it. Calculate its new speed assuming momentum is unchanged.

Firstly you must know the formula for momentum.
Momentum = Mass x Velocity
p = mv

a skateboard of mass 4kg is moving at a steady speed of 2m/s. Calculate its momentum.
p = 4 x 2 = 8 kgms-1

A 1kg bag of sugr is then dropped vertically onto it. Calculate its new speed assuming momentum is unchanged.
p = mv
8 = 5v
v = 1.6 ms-1

 

[Huzza]

thanks!

 

[bamteck]

www.xtremepapers.com/papers/CIE/.../9702_w02_qp_4.pdf

HELP NEEDED for no. 5 :'(

 

[Albert Einstein]

Specimen paper 4 nO 7 second part

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_nos_sp_4.pdf

 

[dao dai]

Question 5 b) and c) please
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_43.pdf

 

[TeKnOzOr]

Question 5 b) and c) please
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_43.pdf

so here it is:
(b)

As you know, capacitance is the reverse of resistance.....with resistor, when finding resistance of 2 resistors in series,we just add them....but in capacitor, when they are in series, we do 1/c= 1/c1 + 1/c2.....etc ....and when in parallel, we just add them normally...!!
So when u'll calculate the total capacitance of this combination here,

for the 2 capacitors in parallel, we get
c=30uf + 30 uf= 60 uf (because they are in parallel)
therefore total capacitance
there is 30uf and 60 uf in series,
so,
1/c =1/30 + 1/60
1/c= 1/20
c=20 uf

 

[TeKnOzOr]

for the (ii),

For a fully capacitor,
Q=CV
Q= 30 x 6 = 180 uC

For parallel capacitor,
Q=CV
V=Q/C
V=180/60
V= 3 V

Total p.d= 6V + 3V = 9V
Therefore, maximum p.d that can be sefely be applied = 9V <---

 

[memorypriest]

www.xtremepapers.com/papers/CIE/.../9702_w02_qp_4.pdf

HELP NEEDED for no. 5 :'(

Q5)
Use Loss in K.E = Gain in G.P.E
(K.E at infinity is zero , G.P.E at infinity is zero)
And you would get the equation.

 

[TeKnOzOr]

(c)
It is given
C= 4700 uf
V1= 18V
V2= 12V

Energy at V1= 1/2 CV1^2
= 1/2 x (4700x10^-6) x (18)^2
= 0.761

Energy at V2= 1/2 CV2^2
= 1/2 x (4700x10^-6) x ( 12)^2
= 0.338

Now, Energy during discharge = o.761- 0.338
which gives................................= 0.42 J <---
Hope u've understand dear...!! If any doubts, plz let me know..!!

 

[Goku]

I posted this on a new thread too,hope im not breaking any rules!!

But please please please help me with this!!

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_41.pdf

thank you in advance!!

Q1 b(ii) and (iii)

Q2 a.

Q3 b(ii)

Q4 (c)


Pleeeeeeeease!! :'(

 

[VelaneDeBeaute]

Goku For Q2(a),
Number of atoms/particles = Number of moles X Avogadro's constant (The old formula we studied in O Levels applies here )
Anyway, we will be applying this formula to find the number of atoms of gas. We have the Avogadro's constant so we only need the number of moles of the gas.
Let's list the data we have.
P = 2.5 X 10^5 Pa
V = 4.5 X 10^3 cm^3 = 4.5 X 10^-3 m^3.
R = (value from data given in booklet) = 8.31 J/ K mol
T = 290 K
Put this into the Ideal gas equation PV = nRT and hence you get the value of n. Place this value of n into this equation Number of atoms/particles = Number of moles X Avogadro's constant and hence you get the answer.

 

[VelaneDeBeaute]

Goku For Q1, part (b)(ii)
First, write down the obvious fact that Force of gravitation is provided by the centripetal force
Fg = Fc
Next, jot down the two formulas of gravitation that seem involved
Fg = GM/x^2 - (1)
g = GM/ R^2 - (2)
From the second formula, we can derieve that
GM = gR^2
Put this value into - (1)
Fg = gR^2 / x^2
Now, jot down the formula for Fc
Fc = xω^2
Now equate Fc and Fg
You'll get
gR^2 / x^2 = xω^2
Hence,
gR = x^3ω^2

 

[Rickster]

Hi, could someone help me explain a couple of multi chioce questions
The link:http://olevel.sourceforge.net/papers/9702/9702_s12_qp_11.pdf
I need help with question 13 and 31.
Thanks

 

[dao dai]

Need help on Q9 b)
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_42.pdf

I understand that voltage gain=-Rf/Ri
but why when the gain = 2, it shows that G = 2 in the equation VOUT= –(4VA + GVB + VC)

 

[Alice123]

october/ november 2004 paper 1 no 13

 

[Silent Hunter]

october/ november 2004 paper 1 no 13

whats the answer ?

B?

 

[Silent Hunter]

weight is causing moments clockwise while the 300N force anticlockwise.... equate them you will get answer weight is the mid of beam