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Lyrx [107]
3 years ago
5

Part B - Permeability of the lipid bilayer Some solutes are able to pass directly through the lipid bilayer of a plasma membrane

, whereas other solutes require a transport protein or other mechanism to cross between the inside and the outside of a cell. The fact that the plasma membrane is permeable to some solutes but not others is what is referred to as selective permeability. Which of the following molecules can cross the lipid bilayer of a membrane directly, without a transport protein or other mechanism
Physics
1 answer:
Helen [10]3 years ago
7 0

This Question is not complete

Complete Question:

Part B - Permeability of the lipid bilayer Some solutes are able to pass directly through the lipid bilayer of a plasma membrane, whereas other solutes require a transport protein or other mechanism to cross between the inside and the outside of a cell. The fact that the plasma membrane is permeable to some solutes but not others is what is referred to as selective permeability. Which of the following molecules can cross the lipid bilayer of a membrane directly, without a transport protein or other mechanism?

a) proteins

b) water

c) ions

d) sucrose

e) lipids

f) carbon dioxide

g) oxygen

Answer:

b) water

e) lipids

f) carbon dioxide

g) oxygen

Explanation:

Diffusion is the means by which molecules like water, lipids, carbon dioxide and oxygen easily move across the lipid bilayer of a membrane directly without the use of a transport protein or any another mechanism.

Diffusion is the way or means through which molecules( such as gases or liquids) move across a membrane from and region of higher concentration to a region of lower concentration.

Before a molecule can pass through the semi permeable membrane without the use of any mechanism or transport protein, it has to be:

a) a small molecule

b) an uncharged molecule

c) an hydrophobic molecule

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Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
natima [27]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • mass of the larger disk = M_2\ =\ 1.6\ kg
  • Radius of the larger disk =R_2\ =\ 5.0\ cm\ =\ 0.05\ m
  • mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

From eqn (1) and (2), we get,

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let 'a_2' be the acceleration of the block in the second case,

From the above expression,

\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2

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3 years ago
A voltage of 12 cos(I000t+45) Vis applied to a circuit in which a resistor of 4 .n, aninductor of L H, and a capacitor of 100 μF
vlabodo [156]

Answer:

0.01 H

Explanation:

V = 12 cos (1000t + 45)

C = 100 micro farad

Let the inductance be L .

When the current and the voltage are in the same phase so it is the condition of resonance.

So capacitive reactance = inductive reactance

Xc = XL

1/ωC = ωL

L = 1 / ω²C

By comparisonV = Vo Cos (ωt + Ф)

ω = 1000 rad/s

L = 1 / (1000 x 1000 x 100 x 10^-6)

L = 1 / 100

L = 0.01H

thus, the inductance of the inductor is 0.01 H.

8 0
3 years ago
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