Physics Mechanical Properties of Fluids
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New answer posted
a month agoContributor-Level 10
The excess pressure inside a soap bubble is given by ΔP = 4T/R, where T is the surface tension and R is the radius.
The pressures are given as P? = 1.01 atm and P? = 1.02 atm. Let the atmospheric pressure be P? = 1 atm.
ΔP? = P? - P? = 1.01 - 1 = 0.01 atm = 4T/R?
ΔP? = P? - P? = 1.02 - 1 = 0.02 atm = 4T/R?
Dividing the two equations: (ΔP? /ΔP? ) = (R? /R? )
0.01 / 0.02 = R? /R? ⇒ R? /R? = 2
The ratio of their volumes is V? /V? = ( (4/3)πR? ³ ) / ( (4/3)πR? ³ ) = (R? /R? )³ = 2³ = 8.
The ratio is 8:1.
New answer posted
a month agoContributor-Level 10
ρdrω²r = ρgdh
ω²∫? rdr = g∫? dh
ω²R²/2 = gh
h = ω²R²/2g = 25ω²/2g
New answer posted
a month agoContributor-Level 10
√2gh = (2r²g/9η) (ρ_t - ρ)
⇒ h = (2/81) (r? g (ρ_t - ρ)²/η²)
⇒ h ∝ r?
After falling through h, the velocity be equal to terminal velocity.
New answer posted
a month agoContributor-Level 10
As we know that Reynolds's number R = ρvD/η
In First case: v? = (0.18*10? ³)/ (π (0.5*10? ²)²*60) = 0.03822 m/s
R = (10³ * 0.03822 * 0.01)/10? ³ = 382.2 < 2000 (Laminar/Steady)
In Second case: v? = (0.48*10? ³)/ (π (0.5*10? ²)²*60) = 0.10191 m/s
R? = (10³ * 0.10191 * 0.01)/10? ³ = 1019.1 < 2000 (Laminar/Steady)
The provided solution has a different calculation for R, leading to a different conclusion.
New answer posted
a month agoContributor-Level 10
Range R = v*t = √ (2gh) * √ (2 (H-h)/g) = 2√ (h (H-h).
For R to be max, dR/dh = 0.
h (H-h) must be max. d/dh (Hh-h²)=H-2h=0.
h=H/2 = 12/2=6m.
New answer posted
a month agoContributor-Level 10
At terminal speed
Mg = Fv = 6πηRv
⇒ V = mg / 6πηR
V = (4/3)πR³ρg / 6πηR
⇒ V = (2/9) * (ρR²g/η)
= (2/9) * (1000 * (0.2 * 10? ³)² * 10) / (1.8 * 10? )
= 4.94 m/s
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