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New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Total oscillation = 1 2 + 1 8 = 5 8

t i m e t a k e n = T 2 + T 1 2 = 7 T 1 2

= 7

New answer posted

5 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

y = A ¯ + B ¯ = A . B ¯

 

 

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

All the charge given to a conducting sphere resides on outer surface.

 

 

 

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Δ E = 1 3 . 6 ( 1 1 2 1 5 2 ) = 1 3 . 6 * 2 4 2 5 e V

h c λ = 1 3 . 6 * 2 4 2 5 e V . . . . . . . . . . ( 1 )

With the help of conservation of linear momentum, we can write

h λ = m H v H h c λ = c m H v H v H = h c λ c m H = 1 3 . 6 * 2 4 2 5 * 1 . 6 * 1 0 1 9 3 * 1 0 8 * 1 . 6 7 * 1 0 2 7 = 4 . 1 7 m / s

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Time period of second pendulum is 2 seconds.                                                                                                                                                                          &nbs

...more

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

R i = ρ l A

R f = ρ ( 1 . 2 5 l ) ( A / 1 . 2 5 ) = ( 1 . 2 5 ) 2 * ρ l A

R f = 1 . 5 6 2 5 * R i

R f R l R l * 1 0 0 = 5 6 . 2 5 %

New answer posted

5 months ago

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P
Payal Gupta

Contributor-Level 10

v=ωk=301=30m/s

v=TμT=μv2=0.135*103102* (30)2=12.15N=1215*102N

New question posted

5 months ago

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New answer posted

5 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

l=12cε0E02=η* (P4πr2) Here η is the efficiency

E0=ηP2πr2cε0=1.25100*10002*3.14*4*3*108*8.85*1012

E0=12.58*3.14*3*8.85*104=13.69Vm=136.9*101V/m

New answer posted

5 months ago

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P
Payal Gupta

Contributor-Level 10

mg – N = ma

N = m (g – a) = 60 * (10 – 1.8) = 60 * 8.2 = 492 N

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