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Physics: Post your doubts here!

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A stationary thoron nucleus (A=200,Z=90)emits an alpha particle with kinetic energy Ealpha(Ea).which is the Kinetic Energy of the recoiling nucleus?
A: Ea/108
B: Ea/110
C: Ea/54
D: Ea/55
E: Ea
Please can anyone solve this?:)
thoron's mass is 220 u can check that in internet i bet that there is some mistake in this question maybe some priting mistake.
mass is inversely proporntional to k.e

mass k.e
4 Ea
216 x
x= (Ea*4)/216
x=Ea/54
 
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thoron's mass is 220 u can check that in internet i bet that there is some mistake in this question maybe some priting mistake.
mass is inversely proporntional to k.e

mass k.e
4 Ea
216 x
x= (Ea*4)/216
x=Ea/54
i have not understand dude Explain briefly Please:confused:
 
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kindly solve my this problem...in detailView attachment 22639
you know that frequency, f = 1/t, where t = period
so,
f = 1/200*10^-6
= 5000 Hz
= 5kHz
look from 0 to 100 microsecond the period is 100 because the wave has only completed half of the full oscillation but the period of a full oscillation is 200 microsecond which is from time 100 to 300 microsecond, that's why t = 200
I couldn't explain properly but I hope u understand what i'm trying to say :)
problem1.jpg
 
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Can someone please explain/provide notes on the

i) superposition of waves.
ii) stationary waves

I can't understand those 2 topics even after constant revision from the books :\

Thanks in advance!
 
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thanks a lot, u mean that the question asks what is the frequency of 1 modulation!!... :)
following is the next part of the same question.Please tell what will be the values?? I know 3 vertical lines will be drawn...but values??

problem2.jpg

you know that frequency, f = 1/t, where t = period
so,
f = 1/200*10^-6
= 5000 Hz
= 5kHz
look from 0 to 100 microsecond the period is 100 because the wave has only completed half of the full oscillation but the period of a full oscillation is 200 microsecond which is from time 100 to 300 microsecond, that's why t = 200
I couldn't explain properly but I hope u understand what i'm trying to say :)
View attachment 22670
 
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Can someone please explain/provide notes on the

i) superposition of waves.
ii) stationary waves

I can't understand those 2 topics even after constant revision from the books :\

Thanks in advance!

Use the CIE book As/ A International Level by Chris Mee ! Its a very good book.
 
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Urgent help needed in MJ 09 21 question 5 part b)... have a mock in 2 days. Physicsss as
 
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wave.PNGwave2.PNG
Urgent help needed in MJ 09 21 question 5 part b)... have a mock in 2 days. Physicsss as
 
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Thanks a ton Soldier313 :D
Is there a chapter wise compilation for Pure Maths 3? If so then pls reply with the link.
Thanks in advance

You're welcome anytime, nop sorry haven't seen a chaptet wise for math, if i do i'll share the link inshaAllah....
 
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A very detailed explanation..much thanks...
could u plz tell me why the carrier wave has high amplitude/voltage when drawn in the graph(C)
welcome :)
uh...i'm not really sure:oops: but i'll try
I think it has a high voltage because it's the main wave that sends information
sorry, that's all I have:oops:
 
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you can look at this as a pully attached with 2 mass, the man is the lighter object while the barrel is heavier object

so here you need the acceleration since its the same rope the tension and acceleration would same for both cases ie for the man and the barrel

first you need to make simultaneous equations relating the man and the the barrels mass with T and a

T-m1g=m1a this is for a mass which is moving upwards
so we use it for the man
T-800=80a ------ eq1

m2g-T=m2a this is the vice versa of the last one
so we use ir for the barrel
1200-T=120a

solving them simultaneously u will get the acceleration
a= 2
now we use kinamatics here
since it is said that "when the head of the man and the bottom of the barrel are at equal level" it is obvious that when they r equal the must have covered half the distance
so the distance we take here is s=9
likely u= 0 , s= 9 , a=2

v^2 -u^2= 2as
solve it and u get v= 6 m/s

i hope it helps but which paper is this
 

Tkp

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View attachment 22678View attachment 22679
Urgent help needed in MJ 09 21 question 5 part b)... have a mock in 2 days. Physicsss as
for the sound at m to be zero that means destructive interference.so for destructive interference the path difference must be be a odd number of multiple wavelength
and for the second conditon if the amplitudes are same so there will be destructive interference as the two waves cancel out each other
so for the first 1 it would be the path difference must be be a odd number of multiple wavelength and 2 nd one amplitude amplitude of both the waves are same at m.
b)v=f lambda.so u put these values and u get 2 value of l(convert the unit in cm).that means the minima would be within the range.now find the path differnce.find the length of s2m by pythagoras theorem which is 128 cm.so the path difference for s2m-s1m is 28 cm.now apply the formula of path difference for destructive interference
28=(n+.5)l put the value of n to o,1,2,3,4.now u will get 2 values which are within the range.so 2 minima would be detected
 
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