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4 months 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
4 months agoContributor-Level 10
The law of radioactive decay is N = N? e? λt, where N is the amount remaining at time t.
Given that at time t, N/N? = 9/16.
So, 9/16 = e? λt
At time t/2, the fraction remaining will be N'/N?
N' = N? e? λ ( t/2 ) = N? (e? λt)¹/²
Substituting the value of e? λt:
N' = N? (9/16)¹/² = N? (3/4)
The fraction remaining is N'/N? = 3/4.
New answer posted
4 months agoContributor-Level 10
For a rotating loop, the induced emf is ε = ε? sin (ωt), where the peak emf ε? = NABω.
Here, Area A = πab. The number of turns N = 1.
So, ε? = B (πab)ω
The average power loss due to Joule heating in a resistor R is given by:
P_avg = (ε_rms)² / R
Where ε_rms = ε? /√2
P_avg = (ε? ²/2) / R = (B (πab)ω)² / (2R) = π²a²b²B²ω² / (2R)
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