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aev [14]
3 years ago
6

Active transport requires energy in order to move molecules ____________ the concentration gradient while simple and facilitated

diffusion require no energy to move molecules __________ the concentration gradient.
Physics
1 answer:
S_A_V [24]3 years ago
4 0

Answer:

Active transport requires energy in order to move molecules <u>against</u> the concentration gradient while simple and facilitated diffusion require no energy to move molecules <u>down</u> the concentration gradient.

Explanation:

Active transport requires energy as the substance move from low concentration to high concentration (<u>substances move against concentration gradient</u>).

For example, Sodium-potassium pump.

Passive transport does not require energy as the substance move from high concentration to low concentration (<u>substances move only down their concentration gradient</u>).

It is of two types: Simple and facilitated transport.

For example, Osmosis.

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White visible light travels from the Sun to Earth. What is the best description of white light?
aliya0001 [1]

Answer:

As investigated by Sir Isaac Newton,the white light contains all colours of light.

This shorten by a term rainbow colours.

6 0
2 years ago
A 2137 kg car moving east at 12.91 m/s collides with a 3264 kg car moving north. The cars stick together and move as a unit afte
IRINA_888 [86]

Answer:

The speed of the 3264 kg car before collision is 6.32 m/s

Explanation:

Let first car, mass m₁, moving east be moving with velocity, v₁ and second car, mass m₂, moving north be moving with velocity, v₂. Let the velocity of the two cars after collision be v₃

In a Collison (whether elastic or inelastic), momentum is always conserved.

Momentum before collision = momentum after collision.

Momentum before collision = m₁v₁ + m₂v₂

v₁ = (12.91î) m/s in vector form, m₁ = 2137 kg

v₂ = ?, v₂ = (v₂j) m/s in vector form, m₂ = 3264 kg

Momentum before collision = (2137)(12.91î) + (3264)(v₂j) = (27590î + 3264v₂j) kgm/s

Momentum after collision = (total mass of the cars after collision) × (velocity of the stuck-together cars after collision)

Total mass of the cars after collision = m₁ + m₂ = 2137 + 3264 = 5401 kg

Velocity after collision, v₃ = 6.38 m/s In the N36.8°E direction, put in vector form,

v₃ = (6.38 cos 36.8°î + 6.38 sin 36.8°j)

v₃ = (5.11î + 3.82j) m/s

Momentum after collision = 5401 (5.11î + 3.82j) = (27590î + 20641.4j) kgm/s

Momentum before collision = momentum after collision.

(27590î + 3264v₂j) = (27590î + 20641.4j)

Comparing components

3264v₂ = 20641.4j

v₂ = 6.32 m/s

Hope this Helps!!!

6 0
3 years ago
On a distance vs. time graph, how do you know when the object is moving away from its starting position?
Ainat [17]
U know by if they are in first place
8 0
2 years ago
Read 2 more answers
Part C: Quantitative Problems when vf is not 0
Alina [70]

Answer:

(a)

\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s

(b)

1120 N

Explanation:

Change in velocity, \triangle v is given by subtracting the initial velocity from the final velocity and expressed as \triangle v= v_f -v_i

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

\triangle v=0-8=-8\ m/s

To find m\triangle v then we substitute 7 kg for m and -8 m/s for \triangle v therefore \triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}

Substituting t with 0.05 s then a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

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3 years ago
Which property of light is not explained by the wave model of light?
Ronch [10]

Blue light can knock electrons off a plate, but red light can't

5 0
3 years ago
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