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4 years ago

The specialized capillary bed responsible for the pressure that drives filtration is the

Physics

1 answer:

Evgen [1.6K]4 years ago

3 0

Answer: GLOMERULUS

Explanation:

The specialized capillary bed responsible for the pressure that drives filtration is the

GLOMERULUS.

The kidney is an organ responsible for the excretion of nitrogenous wastes from the human body and osmoregulation of the blood and body fluids. The structure which is responsible for formation of urine is the nephron( kidney tubule). The GLOMERULUS are found within the Bowman's capsule of the nephrons. It is made up of specialised bundle of capillary beds which are the only capillary beds that are not surrounded by interstitial fluid in the body. In the glomerulus, blood pressure is high because an arteriole enters and exists the capillary beds which is responsible for the pressure that drives filtration.

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Two people are trying to push a large box across a floor. Person 1 pushes with a force of 15 N to the right, while person 2 push

ozzi

Answer:

35 N to the right

Explanation:

When calculating net force when both forces are on the same side you add them when they are going against each other you subtract them.

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3 years ago

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Shtirlitz [24]

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3 years ago

If you go to space at 15000 mph. how long would it take you at the same speed to reach mars

madreJ [45]

Answer:

if your going 15000 mph all you can do is accelerate or go fast unless there is a opposite

Explanation:

6 0

3 years ago

Two 3 kg spherical masses are connected by a 20 m long thread of negligible mass that does not stretch. The masses are charged s

Yuri [45]

Answer:

The charge of each sphere is $q = - 1.84 *10^{-4}C$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The mass of the two sphere is $m = 3 \ kg$

The length of the connection is $l = 20\ m$

The angle between the suspended side is $\theta = 17 ^o$

At Equilibrium the force acting at the horizontal is = 0N and the net force acting at the vertical is zero

This can be represented mathematically as

$\sum F_y = 0N , \ \ and \ \ \sum F_x = 0N$

For $\sum F_y = 0N$

$T cos \theta = mg$

=> $cos \theta = \frac{mg}{T}$

For $\sum F_x = 0N$

$T sin \theta = F$

=> $sin \theta = \frac{F}{T}$

Now $tan \theta = \frac{sin \theta }{cos \theta }$

$= \frac{\frac{F}{T} }{\frac{mg}{T} }$

$= \frac{F}{mg}$

Where F is the electrical force which is mathematically represented as

$F = \frac{kq^2}{r^2}$

Therefore

$tan \theta = \frac{kq^2}{mgr^2}$

Where $r$ is the distance between the two masses and from the diagram it is

$r = 2l sin \theta$

$tan \theta = \frac{kq^2 }{(2l sin \theta )^2 * mg}$

making q the subject of the formula

$q = \sqrt{\frac{mg}{k} * tan \theta (2l sin \theta )^2 }$

Where k is the Coulomb's constant with a value of $k = 9*10^{9} kg \ cdot m^3 s^{-4} A^{-2}$

Substituting values

$q = \sqrt{\frac{3 * 9.8 }{9*10^9} * tan (17) (20 sin 17)^2 }$

$q = - 1.84 *10^{-4}C$

The negative sign is because we are told from the question that they are negatively charged

8 0

3 years ago

Read 2 more answers

A uniform charge density of 509 nC/m3 is distributed throughout a spherical volume of radius 6.03 cm. Consider a cubical Gaussia

Artemon [7]

Explanation:

(a) It is known that relation between charge and volume is as follows.

$q_{enclosed} = \rho V_{cube}$

= $(509 \times 10^{-9} C/m^{3}) \times (0.04 m)^{3}$

= $509 \times 10^{-9} \times 6.4 \times 10^{-5}$

= $3.26 \times 10^{-11} C$

Now, according to Gauss's law

$\phi = \frac{q_{enclosed}}{\epsilon_{o}}$

= $\frac{3.26 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}$

= 3.68 $N m^{2}/C$

Hence, the electric flux through this cubical surface if its edge length is 4.00 cm is 3.68 $N m^{2}/C$ .

(b) Similarly, we will calculate the electric flux when edge length is 16.8 cm as follows.

$q_{enclosed} = \rho V_{cube}$

= $(509 \times 10^{-9} C/m^{3}) \times (0.168 m)^{3}$

= $509 \times 10^{-9} \times 4.74 \times 10^{-3}$

= $2.41 \times 10^{-11} C$

Now, according to Gauss's law

$\phi = \frac{q_{enclosed}}{\epsilon_{o}}$

= $\frac{2.41 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}$

= 2.72 $N m^{2}/C$

Therefore, the electric flux through this cubical surface if its edge length is 4.00 cm is 2.72 $N m^{2}/C$ .

4 0

4 years ago