Physics: Post your doubts here!

[Kanekii]

 

[A*****]

Just the 14th one pls!View attachment 64141

Refer to the examiner report

 

[Tom Sewage]

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)

thank u

 

[Kanekii]

View attachment 64142

Ahh forgot the A^2

 

[A*****]

Is this statement regarding 1% correct?? But the extra 0.1 is 10%, not 1%

Someone plz answer this as well

 

[potatocorn12]

i am confused with both questions ugh

 

[Kanekii]

D right?

 

[A*****]

D right? View attachment 64145

Yes

 

[Kanekii]

View attachment 64144 i am confused with both questions ugh

Q15 C

 

[Mush999]

View 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//

 

[potatocorn12]

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//

Q15View attachment 64146 C

Thank youuu,i suck at physics lol

what about 17 tho

 

[mariam001]

for the feb march paper what is the explanation for question 14 ?

 

[Kanekii]

Is the answer B?

 

[A*****]

Thank youuu,i suck at physics lol

what about 17 tho

 

[A*****]

Is the answer B? View attachment 64147

Yeah!

 

[Mush999]

Thank youuu,i suck at physics lol

what about 17 tho

No problem!
Q17.

Total energy in the fuel
34 × 1.5 × 10^4 = 5.1×10^5kj

And it states that 1.5 × 10^4 is vol used per hour

So per second second 5.1×10^5/60×60 = 141.67kw

So wasted power is 141.67-40= 101.67≈ 102kw

 

[Tom Sewage]

need help, how to solve this question

 

[A*****]

View attachment 64149

need help, how to solve this question

At highest point, vertical comp of velocity is zero and horizontal comp is v cos45= 0.707v
New kinetic energy= 0.707²= 0.50E

 

[Tom Sewage]

View attachment 64144 i am confused with both questions ugh

 

[Ayesha_m01]

ANS is C..?