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NNADVOKAT [17]
2 years ago
7

The energy for _____ transport comes from the gradient itself

Physics
1 answer:
natali 33 [55]2 years ago
4 0

Answer:

the anwser is atp

Explanation:

To move substances against a concentration or an electrochemical gradient the cell must use energy. This energy is harvested from ATP that is generated through cellular metabolism. Active transport mechanisms, collectively called pumps or carrier proteins work against electrochemical gradients.

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Si se aumenta la tentación en una cuerda, ¿que pasa con la velocidad de la onda?​
Leokris [45]

Answer:

El aumento de tensión alarga la longitud de onda, reduce la amplitud, aumenta la frecuencia y, por lo tanto, aumenta la velocidad. (GOOGLE)

4 0
3 years ago
What is the si unit of force ​
Tcecarenko [31]

Explanation:

SI unit of force is <em>Newton</em>

3 0
3 years ago
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The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on
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25 times the average speed

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3 years ago
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A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

W=K_f -K_i

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving,

K_f = K_i + W=66,120+(-36,733)=29387 J

The final kinetic energy of the car can be written as

K_f = \frac{1}{2}mv^2

where

m = 661 kg is its mass

v is its final speed

Solving for v,

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

4 0
4 years ago
A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surfac
7nadin3 [17]

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

i.e.

Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

8 0
3 years ago
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