Class 11th
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New answer posted
4 months agoContributor-Level 9
U and H are temperature dependent
Cp, m – Cv, m = R (for 1 mole of ideal gas)
dU = CvdT
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
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
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
New answer posted
4 months agoContributor-Level 10
Mg + MkV² = MA = -mv (dV/dx)
Vdv = (−) (g + kV²)dx
∫? (Vdv)/ (g + kv²) = ∫? - dx
[ln (g + kV²)/2k]? = -x
ln (g/ (g + ku²) = −2kx
x = (1/2k)ln (1 + ku²/g)
New answer posted
4 months agoContributor-Level 10
Using Bernoulli's equation
P? + (1/2)ρv? ² + ρgh? = P? + (1/2)ρv? ² + ρgh?
For horizontal tube h? = h?
P + (1/2)ρv² = P/2 + (1/2)ρV²
(1/2)ρV² = P/2 + (1/2)ρv²
V = √ (P/ρ + v²)
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