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Can't we do it this way?View attachment 60445
View attachment 60457
idk if im right but i solved it like this
Tcos0 = W
Tsin0 = Fc
Fc = 0.15*6.35^2/1.82 = 3.33N
make T the subject T= 1.47 / cos0
substitute it into Tsin0 = Fc
1.47/cos0 *sin0 = 3.33
tan0 = 3.33/1.47
0 = 66.0
Tcos66 = 1.47
T = 3.6..N
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 degreesCan'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
Oh right! Thanks for correcting me !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
npOh right! Thanks for correcting me !
That angle is correct from the calculations. The answer must have been misprinted.might b bcz the angle of horizontal is correct...66
http://www.cie.org.uk/images/164526-2016-2018-syllabus.pdfView 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
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)Can someone please explain to me what is meant by photon and when it is emitted? Thanks in advance.
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?!?
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) ?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)
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)
Q8bi)On a different note, please solve my probs . I have circled the parts I am facing problems in.
3biv)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
Whats 14.5+7*0.02?3biv)
V = (14.5 + 7)*0.02 = 0.43V <------ Potential = individual sum of electric potentials.
V = W/Q
W = VQ
W = KE
KE = 0.43 * 2 * 1.6 * 10^-19 = 1.38 x 10^-19 J
1 : 1 ratio. Here :~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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