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Lapatulllka [165]
3 years ago
10

Which of the statements is true of facilitated diffusion?

Physics
1 answer:
Alexxandr [17]3 years ago
3 0

Answer:

The correct answer is:

Transport occurs with the help of membrane proteins

Explanation:

Facilitated diffusion is the passive (no energy required) movement of molecules across a cell membrane, with the aid of a membrane protein. energy in form of ATP or GTP is not required since molecules are moving down their concentration gradient.

The need for facilitated diffusion occurs when substances cannot successfully cross the phospholipid bilayer of the cells membrane on their own either due to their polarity or large size. The proteins in facilitated diffusion are tthose that act as carriers and those the form the channels across the membrane, through which the diffusion occurs. Examples of where facilitated diffusion is seen is in the transport of glucose (relatively large molecule) across the cell membranes, with the help of the glucose transporter and the transportation of water molecules across the cell membrane through aquaporins.

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To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c
kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

\lambda = \frac{c}{f}

Substituting,

\lambda = \frac{344}{1250}

\lambda = 0.2752m

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

sin\theta = \frac{m \lambda}{W}

Where,

W is the width

m is the integer

\lambda the wavelenght

Re-arrange the expression,

\theta = sin^{-1} \frac{m\lambda}{W}

For m=1,

\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°

For m=2,

\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°

For m=3,

\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

5 0
3 years ago
What type of material is good at transferring heat?
MatroZZZ [7]
The answer is A

Materials that are good conductors of thermal energy are called thermal conductors. Metals are very good thermal conductors. Materials that are poor conductors of thermal energy are called thermal insulators. Gases such as air and materials such as plastic and wood are thermal insulators
5 0
3 years ago
Amoebas have projections called pseudopods. What are they used for
Dvinal [7]

Answer:

Amoebas have projections called pseudopods

Explanation:

Pseudopodia is the locomatary organ of amoeba. It helps them in movement and transportation.

8 0
2 years ago
In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ????
34kurt

Answer:

The current in the wire is 31.96 A.

Explanation:

The current in the wire can be calculated as follows:

I = \frac{q}{t}

<u>Where</u>:

q: is the electric charge transferred through the surface

t: is the time      

The charge, q, is:

q = n*e

<u>Where</u>:

n: is the number of electrons = 7.93x10²⁰

e: is the electron's charge = 1.6x10⁻¹⁹ C

q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C

Hence, the current in the wire is:

I = \frac{126.88 C}{3.97 s} = 31.96 A

Therefore, the current in the wire is 31.96 A.

I hope it helps you!

3 0
3 years ago
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