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notka56 [123]
4 years ago
8

During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.

0% of its normal value, and the blood pressure difference across the circulatory system has increased by 50.0%. By what factor has the average radii of her blood vessels increased?
Physics
1 answer:
Phoenix [80]4 years ago
7 0

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

Q = \frac{\Delta P \pi r^4}{8\eta l}

Where:

\eta_i = are the viscosities of the concrete before and after the increase

l = Length of the vessel

r_1, R_2 = Radio of the vessel before and after the increase

\Delta P= Change in the pressure

Q_{1,2} = The rates of flow before and after he increase

Our values are given as:

Q_2 = 10Q_1 \rightarrow 10 times her resting rate

\eta_2 = 0.95\eta_1 95% of its normal value

\Delta P_2 = 1.5\Delta P_1 Increase of 50%

Plugging known information to get

Q_1 = \frac{\Delta P \pi r^4}{8\eta l}

Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4

r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}

r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}

r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}

r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}

r_2 = 1.586r_1

Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.

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Sphinxa [80]

Answer:

  • a) See explanation below

  • b) At X.

Explanation:

Please, see the picture attached with the image of the plastic bottle for this question.

<u>(a) Explain why the water could not flow out of the bottle.</u>

What makes the water flow out of the botlle is the force of gravity, whic attracts the water towards the Earth.

When Johnson made a small hole at the bottom of the plastic bottle containing water, the air outside the bottle, which surrounds it and exerts a pressure all over the outer walls of the bottle, exerted a force against the small area of water "over" the hole that is in contact with the air.

Thus, this force of the air pushing upward through the wall opposed the force of gravity pulling downward making the net force zero and the water cannot fall.

<u>(b) To make the water flow out more easily, his teacher suggested making another hole. At which position - X, Y or Z, should he make the 2nd hole in order for the water to flow out the fastest?</u>

You must open the hole at a place where there is not water but air, such that the outer air can enter in the bottle.

That will make that the pressure in the space over the water inside the bottle be equal to the pressure outside.

The pressure of the air above the water will push it downward. Now, the force from the pressure of air inside the water, which is downward, opposes the upward force from the pressure of air around the first hole, and the net force will be downward, making the water flow out more easily.

Thus, the position where he should make the second hole in order for the water to flow fastest is at X.

4 0
3 years ago
Can y'all please help me out with this ?
Inessa05 [86]

Answer:

no where is the main part of the question dude

3 0
3 years ago
Need help solving this question.
MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]

7 0
3 years ago
Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force
xeze [42]

Coulomb's law:

  Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

            = (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C)  / (1.0 m)²

            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
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Why does the sound of a person’s voice sound different underwater than when they are speaking in air?
Tems11 [23]
Sound waves actually travel much faster in water than air, but words and the direction of the noise are distorted.
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