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New answer posted

a year ago

0 Follower 81 Views

V
Vishal Baghel

Contributor-Level 10

-mv cos 60? + 2mu = 0 => v = 4u
½m [v² + u² + 2uv cos 120? ] + ½mu² = mgx sin 60?
=> v² = (8/7)√3gx => ar = (4√3/7)g
∴ t = √ (2L * 7)/ (4√3g)

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

z² * (13.6) (1 - ¼) = 3 * (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) * h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 * 10.2 - Φ) = 1/5.25
=> Φ = 3eV

New answer posted

a year ago

0 Follower 25 Views

V
Vishal Baghel

Contributor-Level 10

Ui = ½C (V/3)² + ½C (V/3)² + ½C (2V/3)²
Uf = ½CV²
Wb = CV²/3
ΔH = [Wb - (Uf - Ui)] = CV²/6 = 0.30 mJ

New answer posted

a year ago

0 Follower 60 Views

V
Vishal Baghel

Contributor-Level 10

Given mg = kL
∴ Iα = (kLθ.L + k (L/2)²θ - mg (L/2)θ)
(mL²/3)α = kL² (3/4)θ (restoring torque)
α = (9k/4m)θ
∴ ω = (3/2)√ (k/m)

New answer posted

a year ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

Is? = 4I?
Is? = I?
∴ I_max/I_min = (√Is? + √Is? )² / (√Is? - √Is? )² = (2+1)²/ (2-1)² = 9/1

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

N = [n + n/10 + n/100 + .]
= n/ (1 - 1/10) = 10/9

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Particles having phase difference of? will move with same speed.

New answer posted

a year ago

0 Follower 20 Views

V
Vishal Baghel

Contributor-Level 10

λ? = h/√2mE? = λ? = h/√2mE?
=> E? = (4/9)E? = 4eV
E? = E? - eV? => V? = 5V

New answer posted

a year ago

0 Follower 27 Views

V
Vishal Baghel

Contributor-Level 10

dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
= ΔL = mω²L²/3πr²Y

New answer posted

a year ago

0 Follower 24 Views

V
Vishal Baghel

Contributor-Level 10

i = 60? , μ = √3
∴ 1 * sin (60? ) = √3 * sin r
=> r = 30?
∴ angle with x-axis = 60?

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