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Ask QuestionPosted by Shuvam Kgajuria 8 years, 1 month ago
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Posted by Sunny Kumar 8 years, 1 month ago
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Varun Banda 8 years, 1 month ago
Proof of Bernoulli's theorem Consider a fluid of negligible viscosity moving with laminar flow.
. Let the velocity, pressure and area of the fluid column be v1, P1 and A1 at Q and v2, P2 and A2 at R.
Let the volume bounded by Q and R move to S and T
where QS = L1, and RT = L2.
If the fluid is incompressible: A1L1 = A2L2
The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume.
Now: Work done = force x distance = p x volume
Net work done per unit volume = P1 - P2 k.e. per unit volume = ½ mv2 = ½ Vρ v2 = ½ρv2 (V = 1 for unit volume)
Therefore: k.e. gained per unit volume = ½ ρ(v22 - v12) p.e. gained per unit volume = ρg(h2 – h1) where h1 and h2 are the heights of Q and R above some reference level.
Therefore: P1 - P2 = ½ ρ(v12 – v22) + ρg(h2 - h1) P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + rgh2 Therefore: P + ½ ρv2 + ρgh is a constant
For a horizontal tube h1 = h2 and so we have: P + ½ ρv2 = a constant
This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa.
No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.
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Varun Banda 8 years, 1 month ago
I'm assuming that when you say series, the springs are attached to each other, forming a long chain.
First off, for this type of question, you need to remember the fact that the force from a spring comes from how long it is stretched multiplied by its constant, k1 or k2. In series, both of these springs would not stretch equally, as they have different spring constants. So, we can come up with 2 equations f1= k1*x1-(1) and f2=k2*x2-(2).
Assuming a force F is applied on this system, the springs will stretch a combined total distance of X.
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Shivam Mishra 8 years, 1 month ago
0Thank You