33@prateek
In time dt let water coming out be volume sdx
mass of water coming out =Ïsdx
momentum =dP=Ïsdxv
F=dP/dt
=Ïsdxv/dt
Ïsvdx/dt=Ïsv2
Be serious enough to answer this .. if u know then only answer . . i don't know the answer .. u gotta give me that with explanation ..
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29 Answers
33
1hey yaar i have a doubt how the force exerted by water coming out is Ïsv2...please explain.....
1in question 3 the rate of production is t2 ... unknown . . .. quantity is minimum at t0 .. not maximum .. solving the differential equation is easy tsk but what is the intial limit .. should i take the initial concentration as zero ..
62dN/dt = t2-λN
when maxima dN/dt=0
so t2=λN(t0)
solve the differential equation above...
dN/dt = t2-λN
now substitute and see :)
62virang!!
See Forces for the spring
1 Force downwards mg
1 force upwards 2mg
Resultant force = mg
Therefore
Total mass of the system = 3m
a = mg/3m
a = g/3 upward for A and downward for B
how do you know that force downward is mg??
you have to first equate it to T
see the whole thing is that T1-T2=ma (spring is massless .. so the tension on both sides of the spring is equal)
Now you have to see that mg-T=ma1
2mg-T=ma2
(you cannot use a1=-a2!!) (because spring is elongable!)
what is the constraint relation??
if the spring is acted by tention T on both sides, then extension is T/k
initially no extension...
iniatilly T is just zero because extension is zero in the spring
hence both accelerate with acceleration of g downwards!
1i dont have solution of any of these .. otherwise i would have figured out myself ... dude ..
1i just remember that force due to water coming out is ... Ïsv2
= 2Ïsgh ...Max Friction is μmg = (5h/V)(ÏV)g = 5Ïgh
now i was confused because in the question it is mentioned that friction in the container is not sufficient to keep the container at rest but s is not given so what to do ... also i noticed that μ is not dimensionless ... [11]
11See Forces for the spring
1 Force downwards mg
1 force upwards 2mg
Resultant force = mg
Therefore
Total mass of the system = 3m
a = mg/3m
a = g/3 upward for A and downward for B
1yaar mujhe first ka (d) lag raha hai.....
i think i'm wrong.....pakka
623rd one is of chain reaction
X - > A - > B
in principle i dont think it in syllabus but it is given almost every where... And I strongly advocate knowing this too....
21initial velocity of fluid comin out : √2gh
Momentum towqards rite = ∂M √2gh
So mom. has to bve equal towardsd left.....
aagge [12] [7]
622nd one is another brilliant question...
the important thing is that initially what will be the velocity of the fluid coming out?
Also apply momentum conservation?
62in the first one also draw the fbd of the spring .. that wil help a lot
it is given that it is massless...
62some one asked this to me just 2 days back
in the first one the most important thing is what is the invariant just after release!