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Artemon [7]
2 years ago
11

What is the eulerian description of fluid motion how does it differ from the lagrangian description?

Physics
1 answer:
Alex_Xolod [135]2 years ago
3 0

Kinematics : Study of motion

Fluid kinematics : study of how fluid flows and how to describe its motion.

There are two ways to describe fluid motion

one is Eulerian, where the variations are described at all fixed stations as a function of time.

the other is Lagrangian, in which one follows all fluid particles and describes the variations around each fluid particle along its trajectory.

<u>DIFFRENCE  BETWEEN  LAGRANGIAN AND EULERIAN:</u>

1.Both Lagrangian and Eulerian describes time variation.

2. Eulerian describes the rate of change in one point of space

Lagrangian descries rate of change of a property of material system.

To know more about the Lagrangian and Eulerian :\brainly.com/question/14944792

#SPJ4

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What is the suitable condition for the superconductivity and high resista (a) Weak phonon-phonon interactions (b) No interaction
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Answer:

(d) A strong electron-phonon interaction

Explanation:

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Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic
VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

m_Agh=\frac{1}{2}m_Av_A^2

where

g=9.8 m/s^2 is the acceleration due to gravity

m_A=0.5 kg is the mass of the block

v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

where:

e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

m_A v_A + m_B v_B = m_A v'_A + m_B v'_B

where

m_B=2 kg

And substituting (1) and all the other values,

m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m

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