All categories

4 years ago

Which of the following is a characteristic of active transport? Check all answers that apply.

Physics

2 answers:

vlada-n [284]4 years ago

5 0

B. It moves substances against a concentration gradient. It requires energy from the cell to.

Hope this helps :)

mojhsa [17]4 years ago

5 0

The following are characteristics of active transport:

B) It moves substances against a concentration gradient.

C) It requires energy from the cell.

<h3>Further Explanation</h3>

Active and Passive Transport

Passive transport is the movement of particles through a membrane which requires no extra energy. It happens spontaneously because there is an imbalance in the amount of solutes or difference in concentration inside and outside the membrane. In order to reach equilibrium or a state where the amount of solute or concentration on both sides is the same, some substances have to cross the membrane.

When substances move from a region of high concentration to low concentration, it is said that the substances are moving down a concentration gradient. This process is spontaneous meaning it does not require energy to occur. The opposite is true for active transport which requires additional energy for it to happen.

Movement of Substances (Passive Transport)

In diffusion, solutes move from a concentrated area to a less concentrated area.

In osmosis, the water molecules move from an area with more water molecules (lower concentration of solutes) to the area with less water molecules (higher concentration of solutes).

<h3 /><h3>LEARN MORE</h3>

semipermeable membrane brainly.com/question/737703
diffusion brainly.com/question/1386629

Keywords: osmosis, diffusion, active transport

You might be interested in

HELP ME ASAP PLEASE!!!!!

solmaris [256]

Answer:

f λ = c = 3.00E8 frequency * wavelength = speed of light

λ = c / f = 3.00E8 / 4.0E14 = 7.5E-7 m

Note 1 angstrom = 1.00E-10 m and 1 milli-micron = 1.00E-9 m

So the wavelength would be 7500 angstroms

The visible spectrum is 4000 A - 7000 A or 400 - 700 milli-microns

3 0

3 years ago

While fishing on a lake, a fisherman notices ripples 0.32 m apart, and they are

Karo-lina-s [1.5K]

the main properties of the main wave propertioes include wavelength amplitude, cruest an trough

5 0

3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

4 years ago

Read 2 more answers

Question 1(Multiple Choice Worth 2 points)

tia_tia [17]

Answer:

Explanation:

the reason for each column is to show the electronic configuration of the elements in that column.

Hope This Helps ▼・ᴥ・▼

3 0

3 years ago

Help with 1 2 and 3 please

geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. $I = \frac{q}{t}$ Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3. So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

$\frac{p1L1}{A1} = \frac{p2L2}{A2}\\$

We are looking for L2 so we can isolate using algebra to get:

$\frac{A2(\frac{P1L1}{A1}) }{P2} = L2$

If you fill in those values you get 0.0205

or 2.05 cm

6 0

3 years ago