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Alexeev081 [22]
3 years ago
11

A parallel plate flow chamber is being designed to study the adhesion of leukocytes to endothelium. The channel width is 2cm; th

e viscosity of the tissue culture liquid medium is 0.0087 g/cm-s. Determine channel heights that can be used to generate wall shear stresses (i.e., ) as high as 20 dynes/cm^2 with flow rates less than 2 ml/s.
Physics
1 answer:
DanielleElmas [232]3 years ago
4 0

Answer:

The channel height is 0.0510 cm.

Explanation:

Given that,

Width w= 2 cm

Dynamic viscosity \mu= 0.0087\ g/cm-s

Discharge flow rate Q= 2 ml/s

Shear stress \tau= 20\ dynes/cm²

We need to calculate the channel heights

Using formula of maximum shear stress

\tau=\dfrca{6Q\mu}{H^2w}

H=\sqrt{\dfrac{6Q\mu}{\tau w}}

Put the value into the formula

H=\sqrt{\dfrac{6\times2\times0.0087}{20\times2}}

H=0.0510\ cm

Hence, The channel height is 0.0510 cm.

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Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south. Find the magni
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Given :

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.

To Find :

The magnitude and direction of

a) A + B .

b) A - B.

Solution :

Let , direction in north is given by +j and east is given by +i .

So , A=-63i and B=63j

Now , A + B is given by :

A+B=-63i+63j

| A+B | = 63\sqrt{2}

Direction of A+B is 45° north of west .

Also , for A-B :

A-B=-63i-63j

|A-B|=63\sqrt{2}

Direction of A-B is 45° south of west .

( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)

Hence , this is the required solution .

4 0
3 years ago
Which of the following holds true for volume if the temperature of a given amount of gas increases and pressure remains constant
svlad2 [7]
D. Volume will increase
Volume and temperature are directly proportional for a gas.
7 0
3 years ago
Read 2 more answers
How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a pu
notka56 [123]

Answer:

Newton's first law

Explanation:

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. Therefore, when the ice is smooth, friction gets lesser, and the force acted on that Puck will be decreased.

6 0
3 years ago
A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?
12345 [234]

The block is made of A) Tin, as its specific heat capacity is 0.225 J/(g^{\circ}C)

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by \Delta T, according to the following equation :

Q=mC_s \Delta T

where  C_s is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

Q = 450 J is the amount of energy supplied to the block

\Delta T = 1^{\circ}C is the change in temperature of the material

Solving the equation for C_s, we can find the specific heat capacity of the unknown sample:

C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

7 0
3 years ago
I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

\frac{1}{2} m {v}^{2}  = mgh

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

\sin(15)  =  \frac{h}{d}  \\ or \: h = d \sin(15)

Plugging this into the energy conservation equation and cancelling m, we get

{v}^{2}  = 2gd \sin(15)

Solving for d,

d =  \frac{ {v}^{2} }{2g \sin(15) }  =  \frac{ {(7.5 \:  \frac{m}{s}) }^{2} }{2(9.8 \:  \frac{m}{ {s}^{2} })(0.259)}   \\ = 11.1 \: m

3 0
3 years ago
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