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seropon [69]
3 years ago
11

If a solution surrounding a cell is hypotonic relative to the inside of the cell, in which direction will water move?

Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0

Answer:

The direction of motion of water molecules will be into the cell.

Explanation:

  • A hypotonic solution is one which has the less concentration of solute in the solvent as compared to the solution on the other side of the semi-permeable membrane.
  • This creates an osmotic pressure gradient across the semi-permeable membrane which is responsible for the flow of water molecules across the membrane until the concentration becomes equal for both the solutions.
  • (A semi-permeable membrane is a sheet or a plane barrier which does not allows the molecules over certain size to pass through it. Here the membrane does not allows the molecules larger in size than that of water molecules to pass through it.)
BabaBlast [244]3 years ago
7 0

Answer:

If the cell is placed in a surrounding solution which is hypotonic in nature.

Then the water from outside of the cell to the inside of the cell. The water will keep on moving from the outside of the cell to the inside of the cell.

The flow of water will take place until the outside environment of the cell and the inside of the cell becomes equal.

The flow of water will take place from the outside of the cell to the inside of the cell.

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The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B=\frac{2\pi*a }{u*I}

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a  = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

I=\frac{2\pi*a*B }{u} Formula(2)

B=0.2*10^{-5}  T = 0.2*10^{-5} \frac{weber}{m^{2} }

a  =8cm=0.08m

u=4*\pi *10^{-7} \frac{Weber}{A*m}

We replace the known information in the formula (2)

I=\frac{2\pi*0.08*0.2*10^{-5}  }{4\pi *10x^{-7} }

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0
3 years ago
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a c
Aneli [31]

Answer

given,

x = (3.9 cm)sin[(9.3 rad/s)πt]

general equation of displacement

x = A sin ω t

A is amplitude

now on comparing

c) Amplitude  =3.9 cm

a) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{9.3\pi}{2\pi}

           f = 4.65 Hz

b) period of motion

        T= \dfrac{1}{f}

        T= \dfrac{1}{4.65}

        T = 0.215 s

d) time when displacement is equal to x= 2.6 cm

x = (3.9 cm)sin[(9.3 rad/s)πt]

2.6 = (3.9 cm)sin[(9.3 rad/s)πt]

sin[(9.3 rad/s)πt] = 0.667

9.3 π t = 0.73

t = 0.025 s

4 0
3 years ago
Read 2 more answers
Compressional stress on rock can cause strong and deep earthquakes, usually at _____.
Serggg [28]
Reverse faults is the most best answer
8 0
4 years ago
Read 2 more answers
Two long parallel wires separated by 4.0 mm each carry a current of 24 A. These two currents are in the same direction. What is
lesya [120]

Answer:

Explanation:

Magnetic field at a a point R distance away

B = μ₀ / 4π X 2I / R where I is current

Magnetic field due to one current

=  10⁻⁷ x 2 x 24 / 1 x 10⁻³

48 x 10⁻⁴ T

Magnetic field due to other current

=  10⁻⁷ x 2 x 24 / 3x 10⁻³

16 x 10⁻⁴ T

Total magnetic field , as they act in opposite direction, is

= (48 - 16 ) x 10⁻⁴

32 x 10⁻⁴ T .

5 0
4 years ago
What is the static friction force and kinetic friction for the image
jasenka [17]

According to the plot, static friction force has a maximum magnitude of around 3.0 N, and kinetic friction has a magnitude of about 1.5 N.

The plot appears to be telling you the force required to get the yellow block moving along the table. If one applies less than 3.0 N of force, the block remains motionless. But as soon as it starts to slide, one need only apply 1.5 N of force to keep it moving (presumably at a constant speed).

8 0
3 years ago
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