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Physics: Post your doubts here!

ZainH

ZainH

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asd said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf
q7
Click to expand...

a) Isotopes are atoms of the same element with a different number of neutrons , but the same number of electrons and protons.
b)i) 92 protons, just look at the periodic table. It's the number near the bottom of the element.
b)ii) Deduct the number of protons from the nucleon number (number at top of elemnt) to get the number of neutrons, 238-92 =146
c)i) For it's mass, you just multiply the uniform mass constant number with your nucleon number, 238*1.66x10^-27 = 3.95x10^-25
c)ii) For density, you just need to divide the mass by volume. This is more maths then physics, you need to know the formula for volume of a sphere which is 4/3pi x r^3. Once you get that just divide the mass by that value.
d) Your density will probably be much more then the value given. Since the actual density is low, you can conclude that most of the atom is empty space or most of it's mass is in the nucleus.
 
asd

asd

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ZainH said:
a) Isotopes are atoms of the same element with a different number of neutrons , but the same number of electrons and protons.
b)i) 92 protons, just look at the periodic table. It's the number near the bottom of the element.
b)ii) Deduct the number of protons from the nucleon number (number at top of elemnt) to get the number of neutrons, 238-92 =146
c)i) For it's mass, you just multiply the uniform mass constant number with your nucleon number, 238*1.66x10^-27 = 3.95x10^-25
c)ii) For density, you just need to divide the mass by volume. This is more maths then physics, you need to know the formula for volume of a sphere which is 4/3pi x r^3. Once you get that just divide the mass by that value.
d) Your density will probably be much more then the value given. Since the actual density is low, you can conclude that most of the atom is empty space or most of it's mass is in the nucleus.
Click to expand...
LOL, I didnt ask for a solution, I was only telling daredevil that this was an example of the question she wanted. :p
Anyways, Good job, bro. :p :ROFLMAO:
 
Tabi Sheikh

Tabi Sheikh

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daredevil when they are in series the current is same in both the resistors therefore Vc/2000=Vr/1500 and Vc+Vr=7.0
and Vc=4.0 Vr=3.0
Then
Vc^2/Rc>Vr^2/Rr
So component C will dissipate more power
 
asd

asd

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Tabi Sheikh said:
daredevil when they are in series the current is same in both the resistors therefore Vc/2000=Vr/1500 and Vc+Vr=7.0
and Vc=4.0 Vr=3.0
Then
Vc^2/Rc>Vr^2/Rr
So component C will dissipate more power
Click to expand...
Dude, how did you calc the resistance 2000 for component C, the graph doesnt even tell that. We only have to figure out that its prob more than that of Resistance R, cause from the graph we know C arleady has a resistance of 1500 at 6 V, so obv at 7 V its going to be higher, well not exactly 2000, or is it going to be 2000?
 
A

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asd said:
Dude, how did you calc the resistance 2000 for component C, the graph doesnt even tell that. We only have to figure out that its prob more than that of Resistance R, cause from the graph we know C arleady has a resistance of 1500 at 6 V, so obv at 7 V its going to be higher, well not exactly 2000, or is it going to be 2000?
Click to expand...
what is the question
 
asd

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Tabi Sheikh said:
daredevil when they are in series the current is same in both the resistors therefore Vc/2000=Vr/1500 and Vc+Vr=7.0
and Vc=4.0 Vr=3.0
Then
Vc^2/Rc>Vr^2/Rr
So component C will dissipate more power
Click to expand...
AHHH, Youre confusing the resistance at Pd of 4v with resistance at pd of 7v. Check your method mate!
 
syed1995

syed1995

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daredevil said:
i'll check out ur question but meanwhile can u tell me that in Q6 bii of the same paper why can't we use the formula v=s/t ??
Click to expand...

That formulae is only when acceleration is 0.. the part before shows that the particle has acceleration s = ut + 1/2at^2 will be used
 
daredevil

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lubna1232 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_23.pdf

guyss Q1 e pleaseee ? and 2a part ii as well. i dont get them :'(
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omigosh.. iwud never have been able to do it without the ms -___- :O

anyways u know that acceleration due togravity is 9.81ms^-2
and acceleration is in the downward direction so it will be 9.81sin15
then apply a formula in which u don't need to put in the final velocity because u don't know it and u can calculate the time simultaneously.
from the three eq of motion the formla we have is:
s=ut + (1/2)at^2

as it is initially at rest so u=0
s=at^2 is the eq u hav
substitute the values and take it from there :)
 
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