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Tju [1.3M]
3 years ago
12

"Water is absorbed in the intestine through which process"? a. simple diffusion b. active transport c. active diffusion d. facil

itated diffusion e. gradient transport
Physics
1 answer:
sdas [7]3 years ago
6 0

Answer: Water is absorbed in the intestine through gradient transport. The correct option is E.

Explanation:

Absorption of water in the mammalian intestines occurs largely on the small intestine through a transport mechanism known as osmosis. Osmosis is the movement of water molecules, through a semi permeable membrane, from the region of lower solute concentration into higher solute concentration to attain an equilibrium.

For water molecules to be absorbed from the lumen into the bloodstream, it's dependent on sodium absorption. This follows the following steps:

-sodium is absorbed by co transport with glucose and amino acids.this aids to move sodium from lumen into the enterocyte.

- Rapid exportation of absorbed sodium through sodium pimps from the enterocyte.

- As sodium is pumped out of cells an osmotic gradient is formed across apical cell membrane, this helps to osmotically drive water across the epithelium. Therefore water is absorbed in the intestine through the osmotic gradient created by sodium absorption. I hope this helps. Thanks.

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A net force of +15 N changes the momentum of an object by +100 kg-m/s. What is the time over which the force is applied? (please
sattari [20]
As momentum / time = force
so; time = 100÷15

so your answer is 6.7 !!
3 0
4 years ago
Why is freshly distilled or deionized water used in this standardization?
sukhopar [10]

Answer:

The amount of carbon dioxide is little in deionized water.

Explanation:

Deionized water is a water with little or no impurities. Impurities are in waters are not able to boil below or above the boiling point of water,and in this case are been retained in the original container.

8 0
3 years ago
Read 2 more answers
Convertir 1200 ms a cs<br> Convertir 0,3 mm a cm.
Vaselesa [24]

Answer:

You can do the reverse unit conversion from cm/s to m/s, or enter any two units below: Metre per second (U.S. spelling: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds.

Explanation:

3 0
3 years ago
A spring with a spring constant value of 125 N/m is compressed 12.2 cm by pushing on it with a 215 g block. When the block is re
allsm [11]

Answer:

v = 2.94 m/s

Explanation:

When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.

Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.

Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means

(1/2)kx^2 = (1/2)mv^2

kx^2 = mv^2

v^2 = (kx^2)/m

v = sqrt((kx^2)/m)

v = x * sqrt(k/m)

v = 0.122 * sqrt(125/0.215)        <--- units converted to m and kg

v = 2.94 m/s

3 0
3 years ago
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

F_p = 6*10^{22}N

F_p = \frac{GMm}{R^2}

F_p = 9*10^{22}N

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r = \frac{R}{2}

Gravitational force is

F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

F = 4F_p

Replacing our values

F= 4(9*10^{22}N)

F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

5 0
3 years ago
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