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irakobra [83]
3 years ago
13

A long, vertical pipe (radius R) filled with an incompressible Newtonian fluid, is initially capped at its lower end. At some in

stant in time, the cap is removed and the fluid begins to flow. a) Simplify the Navier-Stokes equation that describes the axial flow velocity as a function of time and position within the tube. Be sure to include a complete set of boundary/initial conditions. b) Choose appropriate scale factors for the variables that remain in the simplified equations of (a). c) Make the equations and conditions of (a) dimensionless. d) From the dimensional analysis, obtain an estimate of: i) an estimate of the time it takes for the fluid to reach its steady-state velocity ii) a characteristic flow velocity in the tube. e) For an experiment using a specific Newtonian fluid in a pipe of radius 1mm, the flow reaches 95% of its steady-state values in 5 s. If a second Newtonian fluid, having identical density to the first but twice the viscosity was placed in a tube of radius 2 mm, how long will it take before 95% of steady-state is achieved
Physics
1 answer:
marysya [2.9K]3 years ago
7 0

Answer:

98........mmmmmmmmmmmmmmmm

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A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___
34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8·\hat j, in vector form

2) 1.0 km East = 1.0·\hat i, in vector form

3) 1.6 km South = -1.6·\hat j, in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

Final displacement, = 1.0·\hat i + 1.2·\hat j

Where;

Δx = The displacement in the x-direction = 1.0·\hat i

Δy = The displacement in the y-direction = 1.2·\hat j

The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

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Answer:

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Explanation:

Given:

Initial speed of the projectile is, u=60.0\ mm/s

Angle of projection is, \theta=30.0\°

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In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

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The horizontal component of initial velocity is given as:

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Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

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Answer: I would choose options C and D

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