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4 years ago

The process by which water diffuses from a region of greater concentration to a region of lesser concentration is called

Physics

2 answers:

DiKsa [7]4 years ago

8 0

Answer:

osmosis is the correct answer.

Explanation:

The process by which water diffuses from a region of greater concentration to a region of lesser concentration is called osmosis.

Osmosis is a process in which solvent moves through a selective semipermeable membrane into a more concentrated region.

There are three types of Osmotic solutions

isotonic solution where the osmotic pressure is the same across the membrane.

hypotonic solution where the osmotic pressure is lower compared to another solution.

hypertonic solution where the solute concentration is higher as compared to the other solution

Natasha_Volkova [10]4 years ago

5 0

Osmosis is the process by which water diffuses from a region of greater concentration to a region of lesser concentration

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A spring has a length of .200 m when a .300 kg hangs from it,and a length of .750 m when a 1.95 kg hangs from it.what is the for

musickatia [10]

1) 29.5 N/m

2) 0.100 m

Explanation:

The force constant of the spring can be found by using the fact that the force on the spring is proportional to the extension of the spring (Hooke's Law). Therefore, we can write:

$\Delta F= k \Delta x$

where

$\Delta F = F_2 - F_1$ is the change in the force on the spring, where

$F_1 = m_1 g = (0.300)(9.8)=2.94 N$ is the force applied when the hanging mass is

$m_1 = 0.300 kg$

$F_2 = m_2 g = (1.95)(9.8)=19.1 N$ is the force applied when the hanging mass is

$m_2 = 1.95 kg$

$\Delta x=x_2 -x_1$ is the change in extension of the spring, where

$x_1=0.200 m$ is the extension of the spring when the hanging mass is 0.300 kg

$x_2=0.750 m$ is the extension of the spring when the hanging mass is 1.95 kg

Solving for k,

$k=\frac{F_2-F_1}{x_2-x_1}=\frac{19.1-2.94}{0.750-0.200}=29.5 N/m$

When the first mass is hanging on the spring, we have

$F_1 = k (x_1 - x_0)$

where:

$F_1$ is the force applied on the spring (the weight of the hanging mass)

k is the spring constant

$x_1$ is the extension of the spring wrt its natural length

$x_0$ is the natural length of the spring (the unloaded length)

Here we have

$F_1=2.94 N$

k = 29.5 N/m

$x_1=0.200 m$

Solving for $x_0$ , we find:

$x_0 = x_1 - \frac{F_1}{k}=0.200 - \frac{2.94}{29.5}=0.100 m$

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3 years ago

The higher the pressure on a liquid, the higher its boiling point temperature. true false

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True, p1/t1=p2/t2. Pressure is related to temperature at which it boils so pressure does affect the boiling point.

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3 years ago

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3 years ago

A conservative force on a particle moving along the x axis is given by F = (3x^2-2x)i. Which of the following is a potential tha

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2 years ago

The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to Earth's Moon. Their orbital radii

tankabanditka [31]

Answer:

$0.2528$

Explanation:

To calculate the period we need the formula:

$T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$

Where $r$ is the radius of the moon, $G$ is the universal constant of gravitation and $M$ is the mass of mars.

The period of Phobos:

$T_{p}=\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}$

The period of Deimos:

$T_{D}=\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}$

The ratio of the period of Phobos and Deimos:

$\frac{T_{p}}{T_{D}}=\frac{\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}}{\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}}$

$\frac{T_{p}}{T_{D}}=\frac{\sqrt{GM}2\pi r_{p}^{3/2}}{\sqrt{GM}2\pi r_{D}^{3/2}}$

Most terms get canceled and we have:

$\frac{T_{p}}{T_{D}}=\frac{r_{p}^{3/2}}{r_{D}^{3/2}}$

According to the problem

$r_{p}=9,378km\\r_{D}=23,459km$

so the ratio will be:

$\frac{T_{p}}{T_{D}}=\frac{(9,378)^{3/2}}{(23,459)^{3/2}}=\frac{908166.22}{3593058.125}=0.25275$ ≈ $0.2528$

the ratio of the period of revolution of Phobos to that of Deimos is 0.2528

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3 years ago