- Messages
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Refer to the examiner reportJust the 14th one pls!View attachment 64141
27: The formula for path difference is (n-1/2)lamda
n=2, and you can find lamda using x=lamda(D)/a (the double slit formula)
Your lamda will turn out to be 4x10^-9 m
Now just insert the n and lamda into the path difference equation: (2-0.5)(4x10^-7)
Ans will be option C
28: For open tube/pipe, formula for lowest frequency is f=v/2L
v=fx(2L) ---- your open tube equation
For closed tube, formula for lowest frequency is: f=v/4L
v=fy(8L) ----- your closed tube equation
Solve the two equations algebraically now
fx(2L)=fy(8L)
fx=4Fy(D)
Ahh forgot the A^2
Someone plz answer this as wellIs this statement regarding 1% correct?? But the extra 0.1 is 10%, not 1%
YesD right? View attachment 64145
Q15 CView attachment 64144 i am confused with both questions ugh
PE= KE + PEView attachment 64144 i am confused with both questions ugh
PE= KE + PE
it has gained a height and also moving after it slides down!
PE = KE + PE
mgh = 0.5mv^2 + mgh
(m is a constant so cancel it out!)
gh = 0.5v^2 + gh
9.81*h = 0.5*1.4^2 + 9.81*40*10^-2
h= 0.50//
Yeah!Is the answer B? View attachment 64147
Thank youuu,i suck at physics lol
what about 17 tho
At highest point, vertical comp of velocity is zero and horizontal comp is v cos45= 0.707v
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