here probably all wrong correct me...
Q1
A) 40K19 + 0E-1 >>>> 40Ar18 + neutrino
b) weak interaction because proton has converted into a neutron
c)calcualtion... wavelength=8.09*10^-13m
d) 40K19 >>>> 40Ca20 + BETA minus + ANTINEUTRINO
not sure about reasoning my reason was wrong.
Q2))
refracitve
a) i got 39.0
b) n=1.3 something
1.6sin58=nsin 90
c) proabbly wrong but:
red has lower refractive index so its critical angle will be higher than 58 so it does under TIR
but blue has a higher refractive index so it will undergo TIR as its critical angle will belower than 58
nto sure anyone wann say something....
Q3
a)EMF IS y intercept internal is negative gradient
****ed the internal resitance didn't say to take negative..
b) i used the 75mA and said that the battery doesn't provide enough PD would that be right for the first circuit and for the second circuit i did nothing -.-.
c) it said intensity and using the units Intensity = required Power/surface area.
required power= I"R = (75mA)^2 * 6=0.03375W
as efiieciency = 0.04
the power would be= 0.03375/0.04 = 0.84375W
so the Intensity = 0.84375/surface area where area = 0.32cm2 so Intensit= 264Wm^-2
Q4)
ladder beetch question...
a) there is no friction arrow pointing east on the ground so ffriction is not presnt??? not sure about this.
b)NOT SURE ARROW
c)taking moments from the bottom.
4(390cos60) - (8)(Fcos30) = 0
F=65 root 3 = 112.58...=113 N .
d) as the person moves about clockwise moment from the bottom increases so the resultant force at the top has to increase to keep the ladder in position.
not sure about this...
did anyone get similar or same???
Q5)
a) just to lines one with T and one with (M+m)gsin35 i think i forgot the g OH NOOOOOO fook.
b) proving i used some long ass method:
for B: T-Mgsin35=Ma
for A: (M+2m)gsin25-T=(M+2m)a
solve this to equations and then i got the required acceleration... THANK YOU M1
c) I think the question asked about the loaded mass of both A and B, which would be the same mass and velocity therefore the same momentum. (not sure)
THIS IS NOT GUARANTEED
d) i worked out a using the calues M=95 and m=30
and then did 0.75a and used suvat s=ut + ......
and i got t as 4.2
e)4.2 + 12=16.2
30*60/16.2=111 ....not sure about this..
OR
I got 186 blocks
Q6)
A) divide f by root T and then work out propriotnality constant k=50 so they are proprtional..
B)MEW=A*Density so L=0.67m
C) after some large tension the string would break and there would be 0 frequency so you cannot keep using that formula. PROBABLY WRONG
is that right?
Q7)
a) so well we knew f Amax Kinetic Max
so using Vmax=2PIE*f*A
and using EK=1/2 m Vmax " you get m=1.0.... something
and then T=2PIE square root ( m/k) you get k=172.something
so m=1.0,,,
k=170
b) yeh well fook this question....
c) i said the peak amplitude will be lower as ther eis more resistance in oil than in air....
and thta the curve would be wider as it moves more slowly in oil ... not sure about this...