- Messages
- 310
- Reaction score
- 125
- Points
- 53
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I think that it is the lower one wire which is at zero potentialohhh..thank you i never knew about the switch thing
, but what did they mean when they said that a pd of 0v to its input at one switch position
The distance would be 200md = 100m
u = 10m/s
v = 20m/s
v² = u² + 2ad
so
a = (v² - u²)/(2d) = (20² - 10²)/(2*100) = 1.5m/s²
No it's DIs the answer B? View attachment 64138
yes.Is the answer B? View attachment 64138
Wait isint it b??No it's D
t would be 100 as we take inital 10The distance would be 200m
D=200*
For 100no i
t would be 100 as we take inital 10
t passes the marker it's speed is 10 so u=10.For 100
Initial is zero
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%
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