Physics

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New answer posted

2 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

  y = A ¯ + B ¯ = A . B ¯

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

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

2 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

Time period of second pendulum is 2 seconds.

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

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

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Fact based        

 

New answer posted

2 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Bv = B sin 60°

-> B v = 2 . 5 * 1 0 4 * 3 2

E m f = B v * v * l = 2 . 5 * 1 0 4 * 3 2 * 1 8 0 * 5 1 8 * 1 = 1 0 8 . 2 5 * 1 0 3 v o l t s        

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

U = 3PV + 4

n C v T = 3 P V + 4

n * f R T 2 = 3 P V + 4

f = 6 + 8 P V

f > 6 p o l y a t o m i c         

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go throiugh the solution 

  λ = c v = Q 3 W 2 = I 3 T 3 M 2 L 4 T 4 [ λ ] = [ M 2 L 4 T 7 l 3 ]

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  t a n α = v m a x t 1 = a 1

t a n β = v m a x t 2 = a 2

a 1 a 2 = t 2 t 1 t 1 t 2 = a 2 a 1

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