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New answer posted
a year agoContributor-Level 10
ω = 2πf = 1/√ (LC)
L = 1/ (4π²f²C) = 1/ (4π² * 60² * 0.1*10? ) = 70.3 mH
New answer posted
a year ago
Contributor-Level 10
E? (due to A and G) = 2 * (kq/l²)cos (45) = √2 kq/l² (downwards)
E? (due to B and F) = 2 * (kq/l²)cos (45) = √2 kq/l² (towards left)
E? (due to C and H) = 0
E? (due to D and E) = 0
Resultant E = √ (E? ²+E? ²) = √ (2 (kq/l²)²+2 (kq/l²)²) = 2kq/l²
(Solution in the image seems to be different.)
New answer posted
a year agoContributor-Level 10
T = 2π√ (m/k)
0.2 = 2π√ (0.5/k)
k = (0.5) (2π/0.2)² = 50π² ≈ 500
x = A sin (ωt) = 5 sin (2π/T * t)
At t=T/4, x = 5 sin (π/2) = 5cm
PE = ½kx² = ½ * 500 * (0.05)² = 250 * 0.0025 = 0.625 J
New answer posted
a year agoContributor-Level 10
Energy of electron = 3 eV
It forms H atom in n=3 state. Energy released E = 3 - (-13.6/9) = 3 + 1.51 = 4.51 eV.
Photon Energy = 4.51 eV
Threshold energy = hc/λ = 12400eVÅ / 4000Å = 3.1 eV.
kE_max = 4.51 - 3.1 = 1.41 eV
New answer posted
a year agoContributor-Level 10
η_A = W_A/Q? = 1 - T/T?
η_B = W_B/Q? ' = 1 - T? /T
W_A = Q? (1 - T/T? ) = Q? - Q?
W_B = Q? ' (1 - T? /T) = Q? /2 (1 - T? /T)
Given W_A = W_B
Q? (1 - T/T? ) = (Q? /2) (1 - T? /T)
Q? (T? /T) (1-T/T? ) = (Q? /2) (1-T? /T)
(T? /T - 1) = (1/2) (1-T? /T)
2T? /T - 2 = 1 - T? /T
2T? /T + T? /T = 3
T = (2T? +T? )/3
New answer posted
a year agoContributor-Level 10
Degrees of freedom (f) = 3 translational + 3 rotational + (2 * 4) vibrational = 14
γ = 1 + 2/f = 1 + 2/14 = 8/7
W = nR? T/ (γ-1) = (1 * 8.3 * (310-300)/ (8/7 - 1) = (83)/ (1/7) = 581 J
As W is positive, work is done by the gas. The solution says W<0, work done on the gas. This implies? T is negative. The question states temperature rises, so work is done on the gas. W = -582J.
New answer posted
a year ago
Contributor-Level 10
[C] = Q/V = Q/ (W/Q) = Q²/W = (A²T²)/ (M¹L²T? ²) = M? ¹L? ²T? A²
[ε? ] = C/ (4πr²) * (Fr²/q²) = M? ¹L? ³T? A²
[µ? ] = 4πrF/I²l = M¹L¹T? ²A? ²
[E] = F/q = (MLT? ²)/ (AT) = MLT? ³A? ¹
New answer posted
a year agoContributor-Level 10
F = Gm²/ (2R)² = mω²R Given m = 1kg
⇒ ω² = Gm/ (4R³)
⇒ ω = (1/2)√ (G/R³)
New answer posted
a year agoContributor-Level 10
R = R? [1 + α? T]
⇒ 16 = R? [1 + α (15 - T? )]
20 = R? [1 + α (100 - T? )]
Assuming T? = 0°C:
16 = R? (1 + 15α)
20 = R? (1 + 100α)
Dividing the equations:
16/20 = (1+15α)/ (1+100α)
16 + 1600α = 20 + 300α
1300α = 4
α = 4/1300 ≈ 0.003 °C? ¹
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