Answer:
The distance covered by the rocket after fuel ran out is 
Explanation:
Given that the rocket moves with an acceleration 
time 
Since the rocket starts from rest initial velocity 
The distance it travelled within this time is given by 
Velocity at this point is given by 
Given that at this height it runs out of fuel but travels further. Here final velocity 
(maximum height), initial velocity
and time to zero velocity 
Thus it travels 
more after fuel running out. The distance covered during this period is given
Lamina and turbulent flow
Explanation:
mentioning about lamina and turbulent flow we could say that both form in different period of time
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
Since it is restricted at both ends, λ/2 = length of string
λ/2 = 1.5m
λ = 1.5*2 = 3m
