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Physics: Post your doubts here!

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Can't we do it this way?
fc=Tsin0
Mv2/r=Tsino
(0.15)(6.35)^2/1.82= Tsin62.83 (o=wxt)
I get 3.77
but dont u think this formula for theta is used when the length of the entire circumference divided by the radius......and secondly its used for finding the angle in radians ...whereas, we're finding the angle in degrees as trignometric ratios r only applicable for angle in degrees
 
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but dont u think this formula for theta is used when the length of the entire circumference divided by the radius......and secondly its used for finding the angle in radians ...whereas, we're finding the angle in degrees as trignometric ratios r only applicable for angle in degrees
Oh right! Thanks for correcting me !
 
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upload_2016-4-19_19-59-54.png
are these sort of questions included in 2016 syllabus? I've seen them on many papers and they freak me out...I'll be really grateful if someone tells me whether they're included or not.
thanks in advance
 
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View attachment 60472
are these sort of questions included in 2016 syllabus? I've seen them on many papers and they freak me out...I'll be really grateful if someone tells me whether they're included or not.
thanks in advance
http://www.cie.org.uk/images/164526-2016-2018-syllabus.pdf
Page 72 has all the things removed from the syllabus which includes:

29. (a) explain in simple terms the need for remote sensing (non-invasive techniques of diagnosis) in medicine
(o) understand that, in a mobile-phone system, the public switched telephone network (PSTN) is linked to base stations via a cellular exchange
(p) understand the need for an area to be divided into a number of cells, each cell served by a base station
(q) understand the role of the base station and the cellular exchange during the making of a call from a mobile phone handset
(r) recall a simplified block diagram of a mobile phone handset and understand the function of each block
 
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Can someone please explain to me what is meant by photon and when it is emitted? Thanks in advance.
 
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I am so confused about one thing please someone clear it up for me? :3
okay so there's this one question that,
a student suggests that when an ideal gas is heated from 100 degree Celsius to 200 the I.E of the gas is increased, explain if it's correct or not. So, the MS apparently says that no it's not because K.E is directly proportional to the "thermodynamic temperature in Kelvin" thus as ON 12 P41
Them there's kind of the same question in MJ 12 where it says a cylinder contains a gas a constant volume is I'm sunlight so that it's temperature increases from 25 to 35 degree Celsius, does I.E increase? Here in MS it says that it does because of the increase in K.E? Why? Here the temperature is not in kelvin again then how come K.E is increasing?!?
 
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Can someone please explain to me what is meant by photon and when it is emitted? Thanks in advance.
A photon is a packet of energy. Whenever electromagnetic waves propagate energy, there are photons which can be thought of as carrying this energy. So photons are emitted by energy releasing processes (that emit energy in the form of EM waves)
For example, if a nucleus decays and gamma rays are emitted, then photons are emitted. (Photons ARE the gamma)
 
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I am so confused about one thing please someone clear it up for me? :3
okay so there's this one question that,
a student suggests that when an ideal gas is heated from 100 degree Celsius to 200 the I.E of the gas is increased, explain if it's correct or not. So, the MS apparently says that no it's not because K.E is directly proportional to the "thermodynamic temperature in Kelvin" thus as ON 12 P41
Them there's kind of the same question in MJ 12 where it says a cylinder contains a gas a constant volume is I'm sunlight so that it's temperature increases from 25 to 35 degree Celsius, does I.E increase? Here in MS it says that it does because of the increase in K.E? Why? Here the temperature is not in kelvin again then how come K.E is increasing?!?

If temp increases in degrees Celsius, then it definitely increase on any temperature scale you use.
Internal energy increases if work is done on a gas (it's compressed so that it's T is constant, volume decreases, and pressure increases).
Or if heat is supplied to the gas. (volume remains constant, T and Pressure both increase)
If there is anything wrong with what I've said above plz tell me.

Could you link to the paper you're talking about? (the first one)
 
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If temp increases in degrees Celsius, then it definitely increase on any temperature scale you use.
Internal energy increases if work is done on a gas (it's compressed so that it's T is constant, volume decreases, and pressure increases).
Or if heat is supplied to the gas. (volume remains constant, T and Pressure both increase)
If there is anything wrong with what I've said above plz tell me.

Could you link to the paper you're talking about? (the first one)
It is October/November 2012 paper 41. I can't link it because the site is not working. And one other thing, kinetic energy (temperature) of gas molecules only and only increase if the particles collide with more speed right? It has nothing to do with how often the molecules collide, for example in less space(when pressure is increased) ?
 
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If temp increases in degrees Celsius, then it definitely increase on any temperature scale you use.
Internal energy increases if work is done on a gas (it's compressed so that it's T is constant, volume decreases, and pressure increases).
Or if heat is supplied to the gas. (volume remains constant, T and Pressure both increase)
If there is anything wrong with what I've said above plz tell me.

Could you link to the paper you're talking about? (the first one)

On a different note, please solve my probs :). I have circled the parts I am facing problems in.
 

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On a different note, please solve my probs :). I have circled the parts I am facing problems in.
Q8bi)
Number = moles * NA (Avogadro's constant) = (1.2/235) * 6.022 * 10^23 = 3.08 x 10^21
(ii) Let lambda = L
Activity = LN
N = No*e^(-Lt)
N(krypton) = 3.1*10^21*e^(-0.231*3600) = 0 so activity = 0
N(barium) = 3.1*10^21*e^(-6.4*10^(-4)*3600) = 3.1 * 10^20 <------ This many numbers left after one hour.
So Activity = 6.4*10^-4 * 3.1*10^20 = 2 x 10^17
Total Activity = 0 + 2 x 10^17 = 2 x 10^17
 
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Q8bi)
Number = moles * NA (Avogadro's constant) = (1.2/235) * 6.022 * 10^23 = 3.08 x 10^21
(ii) Let lambda = L
Activity = LN
N = No*e^(-Lt)
N(krypton) = 3.1*10^21*e^(-0.231*3600) = 0 so activity = 0
N(barium) = 3.1*10^21*e^(-6.4*10^(-4)*3600) = 3.1 * 10^20 <------ This many numbers left after one hour.
So Activity = 6.4*10^-4 * 3.1*10^20 = 2 x 10^17
Total Activity = 0 + 2 x 10^17 = 2 x 10^17

In first part why did we take Atomic Mass of Uranium instead of Barium? That confused me. In second part why is N0 (initial) of Barium and Krypton taken as same?
 
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