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2 years ago

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7. Plasma from blood (density = 1025 kg/m3) flows along a vertical channel in a steady, incompressible, fully developed laminar

film of thickness h. (i) Simplify the continuity and Navier-Stokes equations to model this flow field. (ii) Obtain expressions for the velocity profile, (iii) the shear stress distribution, (iv) the volume flow rate, and (v) the average velocity. (vi) Relate the plasma film thickness to the volume flow rate per unit depth of surface normal to the flow. (vii) Calculate the volume flow rate for h = 0.5 mm, flowing down a surface b = 2 cm wide.

1 answer:

Goryan [66]2 years ago

6 0

Answer: Q = 8.37×10^-9L/s

Explanation: since the depth of the surface is normal to the flow, ø = 90

Find the attached file for the solution

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7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

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2 years ago

How far will you travel if you walk for 50 seconds at 8 m/s?

Rom4ik [11]

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Explanation:

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2 years ago

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A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

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2 years ago

What will make two bar magnets repel each<br> other?

Kitty [74]

The arrows always start at the magnet's north pole and point towards its south pole. When two like-poles point together, the arrows from the two magnets point in OPPOSITE directions and the field lines cannot join up. So the magnets will push apart

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2 years ago

A traffic light of weight 100 N is supported by two ropes that make 30-degree angle with the horizontal. What is the vertical co

alina1380 [7]

Answer:

$T_{vertical} = 50 N$

Explanation:

given,

traffic light weight = 100 N

angle at which the rope is supported = 30°

vertical component of force = ?

$2 T sin \theta= W$

$2 T sin 30^0= 100$

$T = \dfrac{100}{2 sin 30^0}$

$T = 100\ N$

$T_{vertical} = T sin 30^0$

$T_{vertical} = 100\times \dfrac{1}{2}$

$T_{vertical} = 50 N$

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3 years ago