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drek231 [11]
3 years ago
11

HELP ASAP!!!!!!! PLEASE WILL GIVE BRAINLIEST

Physics
1 answer:
Eva8 [605]3 years ago
7 0

Answer: The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. Not all molecules are free to diffuse. If plasma membrane ruptures or breaks down then molecules of some substances will freely move in and out of the cells. Semipermeable membrane describes a membrane that allows some particles to pass through (by size), whereas the selectively permeable membrane "chooses" what passes through (size is not a factor). Selective permeability is a property of cellular membranes that only allows certain molecules to enter or exit the cell. This is important for the cell to maintain its internal order irrespective of the changes to the environment. For example, water, ions, glucose and carbon dioxide may need to be imported or exported from the cell depending on its metabolic activity. Similarly, signaling molecules may need to enter the cell and proteins may need to be released into the extracellular matrix. The presence of a selectively permeable membrane allows the cell to exercise control over the quantum, timing and rate of movement of these molecules.

Explanation: hope this helps you with your work/assingment! Have a wonderful day/night!

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1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position
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Answer:

The value of the distance is \bf{14.52~cm}.

Explanation:

The velocity of a particle(v) executing SHM is

v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)

where, \omega is the angular frequency, A is the amplitude of the oscillation and x is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., x = 0.

The maximum velocity(\bf{v_{m}}) is

v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Divide equation (1) by equation(2).

\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Given, v = 0.25 v_{m} and A = 15~cm. Substitute these values in equation (3).

&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm

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