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s344n2d4d5 [400]
2 years ago
11

A Newtonian fluid with a viscosity  drains through the space between two large parallel plates as shown in the figure. The gap

distance between the plates is 2b. Obtain relations for the shear stress distribution, shear stress at the walls, velocity profile and volumetric flow rate assuming laminar flow and negligible end effects. You can begin the derivations by considering that vz=vz(x) only and pressure at the inlet and outlet is atmospheric.

Physics
1 answer:
Rina8888 [55]2 years ago
8 0

Answer:

well

Explanation:

basically it's like

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An astronaut is taking a space walk near the shuttle when her safety tether breaks. what should the astronaut do to get back to
GuDViN [60]

get hold of some of her equipment, and throw it away from the craft. she should recoil to the craft ... and  hope ???

7 0
2 years ago
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent
tensa zangetsu [6.8K]

Answer:

(a): The car's relative position to the base of the cliff is x= 32.52m.

(b): The lenght of the car in the ir is tfall= 1.78 sec.

Explanation:

Vo= 0

V= ?

d= 50m

h= 30m

a= 4 m/s²

t= √(2*d/a)

t= 5 sec

V= a*t

V= 20 m/s

Vx= V * cos(24º)

Vx= 18.27 m/s

Vy= V* sin(24º)

Vy= 8.13 m/s

h= Vy*t + g*t²/2

clearing t:

tfall= 1.78 sec (b)

x= Vx * tfall

x= 32.52 m (a)

4 0
3 years ago
Is this right at all
Alja [10]
It is correct! good job :)
7 0
3 years ago
The force of gravity on a 60 kg woman is 588 N. The woman also exerts a gravitational force on
bulgar [2K]

Answer:

the same    588 N

Explanation:

3 0
1 year ago
What can you say about the magnitudes of the forces that the balloons exert on each other?
maxonik [38]

Answer:

F_G=G. \frac{m_1.m_2}{R^2} gravitational force

F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2} electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

F_G=G. \frac{m_1.m_2}{R^2}

where:

G= gravitational constant  =6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}

m_1\ \&\ m_2 are the masses of individual balloons

R= the radial distance between the  center of masses of the balloons.

But when  there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}

where:

\rm \epsilon_0= permittivity\ of\ free\ space

q_1\ \&\ q_2 are the charges on the individual balloons

R = radial distance between the charges.

3 0
3 years ago
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