Physics Mechanical Properties of Fluids
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New question posted
2 months agoNew answer posted
2 months agoContributor-Level 10
v? ²/2 + P? /ρ + gh? = v? ²/2 + P? /ρ + gh?
⇒ v? ²/2 + (P? + ρgh + mg/A)/ρ + g * 0 = v? ²/2 + P? /ρ + g * 0
⇒ v? ²/2 + (ρgh + mg/A)/ρ = v? ²/2 . (1)
Using Continuity equation, we can write
Av? = av? ⇒ v? = (a/A)v?
Putting the value of v? in equation (1), we have
(a²/A²)v? ²/2 + (ρgh + mg/A)/ρ = v? ²/2 ⇒ (A² - a²)/A² * v? ²/2 = (ρgh + mg/A)/ρ
⇒ v? = √ [ (A²/ (A²-a²) * 2 (ρgh + mg/A)/ρ] = 3.09 m/s ≈ 3 m/s
New answer posted
2 months agoContributor-Level 10
P? - P? = 4T/a
P? - P? = 4T/b
P? - P? = 4T (1/a - 1/b)
Also, P? - P? = 4T/r
4T/r = 4T (1/a - 1/b)
1/r = (b-a)/ab
r = ab / (b-a)
New answer posted
2 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
2 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)
New answer posted
2 months agoContributor-Level 10
sx? dg (x? /2) + sx? dg (x? /2)
Uf = (S (x? +x? )/2)gd (x? +x? )/4) x 2
= S (x? + x? )²gd/4
U? -Uf = (Sgd/4) {2x? ² + 2x? ² − (x? + x? )²}
= (Sgd/4) (x? -x? )²
New answer posted
2 months agoContributor-Level 10
ρvg - mg = ma
⇒ pvg/m = g + a
⇒ m = pvg / (g + a)
⇒ 10³ (4/3 π * 10? ) (9.8) / 9.898
⇒ 4.15gm
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