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k0ka [10]
3 years ago
5

Blood takes about 1.65 s to pass through a 2.00 mm long capillary. If the diameter of the capillary is 5.00 μm and the pressure

drop is 2.85 kPa , calculate the viscosity of blood. Assume laminar flow.
Physics
1 answer:
Murrr4er [49]3 years ago
6 0

Answer:

Viscosity of blood will be 918.54\times 10^{-6}Pa-sec

Explanation:

We have given time to pass the capillary = 1.65 sec

Length of capillary L = 2 mm = 0.002 m

Pressure drop \Delta P=2.85kPa=2.85\times 10^3Pa

Diameter d=5\mu m=5\times 10^{-6}m

So radius r=\frac{5\times 10^{-6}}{2}=2.5\times 10^{-6}m

Velocity will be v=\frac{L}{t}=\frac{0.002}{1.65}=1.212\times 10^{-3}m/sec

We know that viscosity is given by \eta =\frac{r^2\Delta P}{8Lv}=\frac{(2.5\times 10^{-6})^2\times 2.85\times 10^3}{8\times 0.002\times 1.212\times 10^{-3}}=918.54\times 10^{-6}Pa-sec

Viscosity of blood will be 918.54\times 10^{-6}Pa-sec

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The answer is C. You must divide your wavelength and your frequency to get your answer.
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3 years ago
A 52.0-kg sandbag falls off a rooftop that is 22.0 m above the ground. The collision between the sandbag and the ground lasts fo
sergiy2304 [10]

Answer:

F = 7,916,955.0N

Explanation:

According to newtons second law

Force = mass * acceleration

Given

mass = 52.0kg

distance S = 22.0m

time t = 17.0 ms = 0.017s

We need to get the acceleration first using the formula;

S = ut+ 1/2at²

22 = 0 + 1/2 a(0.017²)

22 = 0.0001445a

a = 22/0.0001445

a = 152,249.13m/s²

The magnitude of the average force exerted will be;

F = ma

F = 52 * 152,249.13

F = 7,916,955.0N

4 0
3 years ago
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
3 years ago
Discuss how friction is reduced in oce skating
belka [17]

Answer:

below

Explanation:

Ice melts, meaning it has a watery layer upon its surface. This allows things to be moving like they are on a liquid but it has the solidity of a solid. The thin metal of the ice skates also decrease the surface area meaning it exerts more force but in turn, it allows you to move faster and further reduces friction.

7 0
2 years ago
How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'
Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity (K.E.=\frac{1}{2}mv^2). If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height(P.E.=mgh). If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.


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3 years ago
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