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nikitadnepr [17]
3 years ago
7

When the blood flows through the capillary bed, most of the plasma will return to the heart through the venules. What happens to

the plasma that does not travel to the venules? It is removed by the lymphatic vessels. It is lost as heat. It returns to the arterioles. It stays in the capillary bed.
Physics
2 answers:
Bas_tet [7]3 years ago
8 0
The answer is the 2nd sentence.
Vlad [161]3 years ago
3 0

The correct answer is option A , It is removed by the lymphatic vessels

There is loss of watery plasma in the body that creates a hyperosmotic solution within the capillaries and venules. This loss causes the blood plasma (nearly 85%) to move in backward direction into the capillaries near the venules . The remaining 15% blood plasma, move out into the lymphatic vessel through the interstitial fluid. This lymphatic fluid is similar in composition to the interstitial fluid and passes through lymph nodes before it returns to the heart via the vena cava.

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this event is the result of a sudden vertical offset in the ocean floor and is most often the result of plate movement on the oc
MissTica

Answer: Tsunami.

Explanation :

Tsunami is the result of a sudden vertical offset in the ocean floor and is most often the result of plate movement on the ocean floor.

Tsunami is caused due to the displacement of a large volume of water like in an ocean. It consists of a series of waves. It destroys coastlines and coastal settlements. It is also known as a tidal wave.

So, the correct option is (b) Tsunami.

6 0
3 years ago
In your opinion, how can we increase a student’s understanding and respect for science?
astraxan [27]

Answer: i would do something that is really easy and i will do test 2 where there aint a word problems and i am going 2 put the work 2 where it is 2 there standard i aint going 2 do sum over there standards of science.

Explanation:

6 0
3 years ago
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Define resistance and describe what would happen to a light bulb if the voltage increased but the resistance stayed the same. (h
PtichkaEL [24]

The light bulb would glow brighter.

<h3>What is Resistance?</h3>

a force that works against a body's direction of motion and seeks to stop or slow down motion, such as friction. a measure of how much a material prevents an electric current from flowing as a result of a voltage.

What is the law of resistance?

Resistance and Ohm's Law. According to Ohm's law, the resistance of the circuit and the current or energy travelling through the resistance are both exactly proportional to the voltage or potential difference between two places.

The current would grow since it is exactly proportionate to the voltage, increasing the light bulb's brilliance, or simply making it brighter.

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6 0
1 year ago
A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe
Sindrei [870]

Answer:

The magnitude of the magnetic field is 9.3\times 10^{-5}\ T.

Explanation:

Given that,

Charge, q=-8.5\ \mu C=-8.5\times 10^{-6}\ C

Speed of the charged particle, v=9\times 10^6\ m/s

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, F=5.9\times 10^{-3}\ N

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

F=qvB\ \sin\theta

B is the magnetic field.

B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T

So, the magnitude of the magnetic field is 9.3\times 10^{-5}\ T. Hence, this is the required solution.

4 0
3 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
3 years ago
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