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New answer posted
4 months agoContributor-Level 10
i? = 40/100 = 2/5
i? = 40/200 = 1/5
V? – V? = 40i? = 40 * 2/5
V? – V? = 16
V? – Vd = 90i? = 90/5 = 18
V? – Vd = 18 – 16 = 2 volt
New answer posted
4 months agoContributor-Level 10
T = constant
P = constant
PV = nRT
PdV = nRdT
PdV + VdP = 0
ΔV = nRΔT/P
dV = (-)VdP/P
|ΔV| = V (ΔP/P)
V/P ΔP = nRΔT/P
ΔT = V/nR ΔP
C = V/nR T/P = 300/2 = 150
New answer posted
4 months agoContributor-Level 10
P = 1/f = (N/100)D
2x + 40 = 100
x = 30 cm
100 – x = 70 cm
1/v - 1/u = 1/f
1/70 - 1/ (-30) = 1/f
1/f = 1/70 + 1/30 = (3+7)/210 = 1/21
f = 21 cm = 0.21M
Power = 1/f = 1/0.21; D = (100)/21
N/100 D = 100/21
N = 10000/21 = 476.19
N = 476
New answer posted
4 months agoContributor-Level 10
dm = λdx = λ? (1 + x/L)dx
M = ∫? λ? (1 + x/L)dx = λ? [L + L²/2L] = 3λ? L/2
dI = dmx² = λ? (1 + x/L)dx * x²
I = λ? ∫? (x² + x³/L)dx = λ? [L³/3 + L? /4L]
I = (7λ? L³)/12 = (7/12) * (2M/3L) * L³ = (7/18)ML²
New answer posted
4 months agoContributor-Level 10
de-Broglie wavelength
λ = h/mv = h / √ (3kT/m * m) = h / √ (3mkT)
λ = (6.63 * 10? ³? ) / √ (3 * 4.64 * 10? ²? * 1.38 * 10? ²³ * 400)
λ = (6.63 / 2.77) * 10? ¹¹ = 2.39 * 10? ¹¹ m ≈ 0.24Å
New answer posted
4 months agoContributor-Level 10
B (ΔV/V) = ΔP
ΔV/V = ΔP/B = (4 x 10? )/ (8 x 10¹? ) = 1/20
V = l³
dV = 3l² dl
dV/V = 3l²dl/l³ = 3dl/l
ΔV/V = 3 (Δl/l); 1/20 = 3 (Δl/l); ΔV/V = 1/60
% (Δl/l) = 100/60 = 1.67%
New answer posted
4 months agoContributor-Level 10
ΔQ = heat supplied
ΔW = work done
ΔU = change in internal energy
(i) adiabatic (B) Δθ = 0
(ii) isothermal (D) ΔU = 0
(iii) isochoric (A) ΔW = 0
(iv) isobaric (C) ΔU ≠ 0, ΔW ≠ 0, ΔQ ≠ 0
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