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Anna71 [15]
3 years ago
7

In general it is expected that ________. A) osmotic pressure will remain relatively consistent throughout the capillary bed B) o

smotic pressure will be lower in the arteriole end of the capillary bed compared to the venous end C) osmotic pressure will be higher in the arteriole end of the capillary bed compared to the venous end D) hydrostatic pressure will remain constant throughout the capillary bed
Physics
1 answer:
Tanya [424]3 years ago
7 0

Answer:

B. OSMOTIC PRESSURE WILL BE LOWER IN THE ARTERIOLE END OF THE CAPPILLARY BED COMPARED TO THE VENOUS END.

Explanation:

This is true for filtration to take place in the cappillary bed. Osmotic pressure is the net pressure that drives movement of fluid from the interstitial fluid back into the capillaries. Osmotic pressure increase favors reabsorption as water moves from region of higher water concentration in the interstitial fluid to the lower region of water concentration in the capillaries.

At the ends of a capillary bed, the difference in the hydrostatic and osmotic pressures provides a net filtration or reabsorption ratio. At the arteriole end of the capillary bed, hydrostatic pressure is greater than the osmotic pressure allowing movements of fluid to the interstitial fluid (filtration) while as the blood moves to the venous end, the osmotic pressure becomes greater than than hydrostatic pressure.

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The density of lead is 30.2g/cm^3.what is the value in kilograms per meter cube?
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If a short-wave radio station broadcasts on a frequency of 9.065 megahertz (MHz), what is the wavelength of
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If the radio wave is on an FM station, these are in Megahertz. A megahertz is one ... Typical radio wave frequencies are about 88~108 MHz .

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Read 2 more answers
Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center
kiruha [24]

Answer:

The angular  velocity is w_f =  1.531 \ rad/ s

Explanation:

From the question we are told that

     The mass of each astronauts is  m =  50 \ kg

      The initial  distance between the two  astronauts  d_i  =  7 \  m

Generally the radius is mathematically represented as r_i  =  \frac{d_i}{2} = \frac{7}{2}  =  3.5 \  m

      The initial  angular velocity is  w_1 = 0.5 \  rad /s

       The  distance between the two astronauts after the rope is pulled is d_f =  4 \  m

Generally the radius is mathematically represented as r_f  =  \frac{d_f}{2} = \frac{4}{2}  =  2\  m

Generally from the law of angular momentum conservation we have that

           I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}

Here I_{k_1 } is the initial moment of inertia of the first astronauts which is equal to I_{p_1} the initial moment of inertia of the second astronauts  So

      I_{k_1} = I_{p_1 } =  m *  r_i^2

Also   w_{k_1 } is the initial angular velocity of the first astronauts which is equal to w_{p_1} the initial angular velocity of the second astronauts  So

      w_{k_1} =w_{p_1 } = w_1

Here I_{k_2 } is the final moment of inertia of the first astronauts which is equal to I_{p_2} the final moment of inertia of the second astronauts  So

      I_{k_2} = I_{p_2} =  m *  r_f^2

Also   w_{k_2 } is the final angular velocity of the first astronauts which is equal to w_{p_2} the  final angular velocity of the second astronauts  So

      w_{k_2} =w_{p_2 } = w_2

So

      mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2

=>   2 mr_i^2 w_1 = 2 mr_f^2 w_2

=>   w_f =  \frac{2 * m * r_i^2 w_1}{2 * m *  r_f^2 }

=>    w_f =  \frac{3.5^2 *  0.5}{  2^2 }

=>   w_f =  1.531 \ rad/ s

       

3 0
2 years ago
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