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Physics: Post your doubts here!

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Your informations seems perfect but I guess you messed the formula up! If its E=1/2 mv^2 then should V=√2E(divided by)M. Because M is multiplying with V and would go and divide there?
can u please point out my mistake please
 
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can u please point out my mistake please

Would you mind being a little more specific? I dont know which answer you are talking about! Sorry :(
But perhaps not many here would have any mistakes nonetheless I would try only if you identify the question. Thanks :)
 
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A beam of red light has twice the intensity of another beam of the same colour. Calculate the ratio of the amplitude of the waves.
 
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The diagram shows five energy levels in a mercury atom:the ground state and four excited states.Calculate the longest wavelength photon that could be emitted when this atom is excited by collision with a 8eV electron.

________________________________ 0

________________________________-1.1eV


________________________________-3.7eV

________________________________-5.5eV



________________________________-10.4 eV ground state.
 
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Hi friends I seriously need help for the question below It's june 2007 paper 2 number 3 c (ii)
View attachment 21237
PLEASE HELP DEAR FRIENDS
as water is going to expand on evaporating so
lets establish a direct relation
g cm^3
1/1600 1
1 x

we got x as 1600 in the first part right. 1600cm^3 in 1 g
now we want mean separation between particles which is to be in cm not cm^3 so cube root this and we get our answer as 11.7
 
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To make a potentiometer, a driver cell of 4.0 V is connected across a 1.00 m length of resistance wire.
a. a) What is the potential difference across 1 cm length of wire? What length of wire has p.d. of 1.0 V across it?
b. b) A cell of unknown e.m.f E is connected to the potentiometer and the balance point is found at a distance of 37.0 cm from the end of the wire to which the galvanometer is connected. Estimate the value of E. Explain why this can only be an estimate.
c. c) A standard cell of e.m.f 1.230 V gives a balance length of 31.2 cm Use this value to obtain a more accurate value for E.
this is from which past paper??
 
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Guys can someone please help me with the physics P5 paper
I cant understand how to calculate the absolute uncertainties of different values in the table in Q2's of the paper :(
Any help would be grateful
Please reply ASAP!
 
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I need help with the following question ASAP
The density of a material of rectangular block was dteremined by measuring the mass and the linear dimensions of the block. Below are the results obtained with their uncertainities
m=25+/-0.1g
l=5.00+/-0.01cm
b=2.00+/-0.01cm
h=1.00+/-0.01
The density was calculated to be 2.5gcm^-3. What was the uncertainity with this result?
 
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I need help with the following question ASAP
The density of a material of rectangular block was dteremined by measuring the mass and the linear dimensions of the block. Below are the results obtained with their uncertainities
m=25+/-0.1g
l=5.00+/-0.01cm
b=2.00+/-0.01cm
h=1.00+/-0.01
The density was calculated to be 2.5gcm^-3. What was the uncertainity with this result?

density = mass/vol

volume = l x b x h
Δv/v = Δl/l + Δb/b + Δh/h
Δv/v = 0.01/5 + 0.01/2 + 0.01/1
Δv/v = 17/1000

Δp/p = Δm/m + Δv/v
Δp/2.5 = 0.1/25 + 17/1000
Δp = 0.05
 
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GUYS PLZ I NEED HELP WITH THIS PAPER 4 QUESTION ASAP.SOMEONE PLZ REPLY.

The diagram shows five energy levels in a mercury atom:the ground state and four excited states.Calculate the longest wavelength photon that could be emitted when this atom is excited by collision with a 8eV electron.

________________________________ 0

________________________________-1.1eV


________________________________-3.7eV

________________________________-5.5eV



________________________________-10.4 eV ground state.
 
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could someone please tell me if its okay to break both the x-axis and the y-axis for the graphs in P3 of physics? Or are we supposed to break the x-axis?
a quick reply would be highly appreciated! thanks. :)
 
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hi,i need physic 0625 may/june 2002, paper 3,it would be nice if someone has it,,,thanks...
 
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