Class 11th
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New answer posted
7 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
7 months agoContributor-Level 10
By Conservation of Angular Momentum (COAM) about O, just before and after the collision:
Initial angular momentum L? = mv?
Final angular momentum L? = Iω? = (I_rod + I_block)ω? = (M? ²/3 + m? ²)ω?
L? = L?
mv? = (M? ²/3 + m? ²)ω?
ω? = mv / (M? /3 + m? ) = (16) / (21/3 + 1*1) = 6 / (5/3) = 18/5 rad/s
By Conservation of Total Mechanical Energy (COTME) after collision until it comes to rest:
Initial Energy (at the lowest point) = Rotational K.E. = (1/2)Iω? ²
Final Energy (at the highest point) = Potential Energy = mgh_m + Mg h_M
mgh_m = mg (? -? cosθ)
Mg h_M = Mg (? /2 - (? /2)cosθ)
(1/2) * (M? ²/3 + m? ²) * ω? ² = (mg + Mg/2) *
New answer posted
7 months agoContributor-Level 9
3+2√-54=3+6√6i= (3+√6i)².
The difference is ± (3+√6i)? (3-√6i).
Possible values are 2√6i, -2√6i, 6, -6.
Imaginary part is ±2√6.
New answer posted
7 months agoContributor-Level 10
KE_max = hc/λ - φ = 6.63e-19 - 4.41e-19 = 2.22e-19 = 222e-21 J.
New question posted
7 months ago
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