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New answer posted
a year agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-Level 10
While approaching: v = v? (c / (c - vcosθ)
While receding: v = v? (c / (c + vcosθ)
New answer posted
a year agoContributor-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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