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New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

ΔW = area under P — t graph
= (1/2) (4 + 6) * 7 = 35 J
Work done = change in KE ⇒ 35 = (1/2) * 2 * v² - (1/2) * 2 * (1)² ⇒ v = 6 m/s

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

  π 2 a 2 F 4 L = π 2 * 1 π 2 1 0 4 * 1 = 2 . 5 J

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

U = kr² ⇒ F = -dU/dr = -2kr ; 2kr = mv²/r ⇒ v = √ (2k/m) or T = 2πr/v = 2π√ (m/2k)

New answer posted

2 months ago

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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

2 months ago

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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

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2 months ago

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New answer posted

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A
alok kumar singh

Contributor-Level 10

n (7800) = (n + 1) 5200

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2 months ago

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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

2 months ago

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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

2 months ago

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V
Vishal Baghel

Contributor-Level 10

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

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