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New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  Δ L = L 0 α Δ T

= 12 * 11 * 10-6 * 30

= 3960 * 10-6 m

New answer posted

a year ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Photon energy  = h c λ = 1240 e V . n m 500 n m = 2.48 e V = 2.48 * 1.6 * 10 - 19 J = 4 * 10 - 19 J

No. of photons emitted per sec =   power     photon Energy   = 100 4 * 10 - 19 s - 1 = 2.5 * 10 20 s - 1

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

n (7800) = (n + 1) 5200

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

L = Iω
I = (mL²/3) + (mL²/3) + [m (√2L)²/12] + m (L/√2)² = mL² [2/3 + 1/6 + 1/2] = (4mL²/3); L = (4/3)mL²ω

New answer posted

a year ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

  I = 1 2 − 0 . 3 5 * 1 0 3 = 2 . 3 4 m A

V 0 = I R = ( 2 . 3 4 * 1 1 − 3 ) ( 5 * 1 0 3 ) = 1 1 . 7 V  

            

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

v? = 2î F? = -2?
v? = 2? F? = -2î
⇒ B? is along -k? Hence v? = 2k? ⇒ F? = 0

New answer posted

a year ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

  τ = ∫ F . r = ∫ 0 R η ( 2 π r     d r ) ( r w t ) . r  

  = 2 π η W R 4 4 t = 2 * 3 . 1 4 * 1 * 8 * 1 0 − 4 4 * 2 * 1 0 − 3          f

  = 0 . 6 2 5 N − m ? 0 . 6 3 N − m            

New answer posted

a year ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Gain in K.E. =  Loss in P.E.

  K p = e V for proton

K α = 2 e V  for  α - particle

Again, Loss in K.E. =   Work against field

K p = e E ⋅ S 0  for proton

⇒ e V = e E S 0

∴ S 0 = V E

K α = 2 e E . S  for α -  particle

⇒ 2 e V = 2 e E S

∴ S = V E = S 0

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Vibrational energy of a non-rigid gas molecule is K? T so, total energy = (5/2)K? T + K? T = (7/2)K? T
∴ C? = (7/2)R

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

tan 60° = (2kλ? /y? )/ (2kλ? /x? )


(√3λ? /x? ) = (λ? /y? ); √3λ? y? = λ? x?

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