Physics

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New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

τ = RC = 10µS
For 0

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

I = m (l/√2)²*2 + m (√2l)² = 3ml²
L=Iω = 3ml²ω

New answer posted

a year ago

0 Follower 15 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

Net field along AB at O must be zero.
E? cosα = E? sinα
(kQ? /x? ²) (x? /AB) = (kQ? /x? ²) (x? /AB)
Q? /Q? = x? ³/x? ³

New answer posted

a year ago

0 Follower 2 Views

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Vishal Baghel

Contributor-Level 10

Q = (1/R)√ (L/C) = (1/100)√ (80e-3/2e-6) = 2

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

F = -dU/dr = - [-12A/r¹³ + 6B/r? ]
F=0 ⇒ r= (2A/B)¹/?
U (at r= (2B/A)¹/? ) = -A²/4B

New answer posted

a year ago

0 Follower 14 Views

V
Vishal Baghel

Contributor-Level 10

Φ? = π/2 - (2π/λ)x = π/2 - (2π/20)5 = 0
Φ_B = π/2
Φ_C = π/2 + (2π/λ)x = π/2 + π/2 = π
I_A = 4I? ; I_B = 2I? ; I_C = 0

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

While approaching: v = v? (c / (c - vcosθ)
While receding: v = v? (c / (c + vcosθ)

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

L.C. = 0.5 mm / 50 = 10? ² mm = 10? m = 10µm

New answer posted

a year ago

0 Follower 3 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

Surprisingly, the viscosity of a dilute gas behaves exactly opposite to what you might expect for liquids. Liquid viscosity generally decreases as temperature is lowered. The viscosity of a dilute gas increases as and when you raise its temperature. This counter-intuitive behaviour was clearly established experimentally and is explained by the kinetic theory.

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