Ask question

Ask question

All categories

4 years ago

14

Velocity profile of a moving fluid over a flat (fixed) plate is given as: ???????? = from the fixed plate (yy = 0) in mm, u is t

he velocity in xx direction in mm/????????. Calculate the shear stress on the fluid at yy = 0.15 m if the viscosity of the fluid 8.6 poise (1 Poise= 1 ????????yy????????????????.????????) cccc2

1 answer:

schepotkina [342]4 years ago

8 0

Answer: The question has some details missing which is the velocity profile .The velocity profile is given as u = (8y - 0.3y^2) mm/s

the shear stress on the fluid at y = 0.15 m = 6.80 x 10^-3N/m^2

Explanation:

The detailed steps is the use of the relationship between the shear stress, viscosity and the velocity profile. appropriate differentiation and substitution were made to calculate the shear stress as is shown in the attachment.

You might be interested in

Rank the following in terms of increasing momentum:

professor190 [17]

It’s b because if you’re running at 5 miles per second being a 100kg weight is the fastest

5 0

4 years ago

Read 2 more answers

When an 8-V battery is connected to a resistor, 2 A of current flows in the resistor. What is the resistor's value? A. 2 ohms B.

SVEN [57.7K]

Its B 4ohM
As v =I /R
R=V/I

6 0

4 years ago

ASAP pls need help will give 25 points per person

faltersainse [42]

Explanation:

1. inertia

2.unbalanced?

3.inertia

4.more

5.inertia

6.inertia

7.unbalanced

8 0

3 years ago

Which of the following is a good step in learning how to manage time?

Luden [163]

I think is c because your seeing how your using you’re time.

8 0

3 years ago

Read 2 more answers

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago