Class 11th
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New answer posted
2 months agoContributor-Level 10
Using conservation of linear momentum, we can write
P? = P? ⇒ mv = (M+m)V
Using conservation of Mechanical energy, we can write
½ (M+m)V² = (M+m)gh ⇒ V = √2gh
⇒ v = (M+m)/m)√2gh
= (6/0.01)√ (2*9.8*0.098) = 600 * 1.386 = 831.4 m/s
New answer posted
2 months 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
2 months agoContributor-Level 10
λ = kT / (√2πd²P)
= (1.38*10? ²³ * 300) / (√2 * 3.14 * (0.3*10? )² * 1.01*10? )
≈ 102 nm
New answer posted
2 months agoContributor-Level 10
using hook's law:
σ = Yε ⇒ f/A = Y (x/l) ⇒ Y = fl/ (xA) = fl/ (xπr²)
Using error analysis formula:
ΔY/Y = Δf/f + Δl/l + Δx/x + 2Δr/r
%error in Y = [ (Δm/m) + (Δl/l) + (Δx/x) + 2 (Δr/r) ] * 100
= [ (1/1000) + (1/1000) + (0.001/0.5) + 2 (0.001/0.2) ] * 100
= [ 0.001 + 0.001 + 0.002 + 0.01 ] * 100 = 1.4%
New answer posted
2 months agoContributor-Level 10
v = 0.5t² i + 3t j + 9 k m/s ⇒ a = dv/dt = (ti + 3j) m/s²
At t=2sec, v = 2i + 6j + 9k m/s and a = (2i + 3j) m/s²
The direction of acceleration of mosquito after 2s is given by the angle θ with the y-axis, where tanθ = a? /a? = 2/3.
So, the direction is tan? ¹ (2/3) from the y-axis.
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