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notka56 [123]
3 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]3 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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OLga [1]

Answer:

<h2>velocity = 12.73 km/hr.</h2><h2 />

Explanation:

velocity = distance / time

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                  2.2 hr

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5 0
2 years ago
You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

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3 0
3 years ago
What type of wave has particles push together and pull apart?
wariber [46]

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5 0
3 years ago
Which of the following statements about iron filings placed upon glass resting on top of a bar magnet is false? A. The direction
IceJOKER [234]
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Answer:

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mass = 7.65 kg

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solution

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Fa =  weight × kinetic coefficient of friction

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Fa = 2.25 N

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