We have starin which is 0.4 and area is 0.05 what is sigma
1 answer:
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Answer: W = J
Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.
The work to transport an ion from a lower potential side to a higher potential side is calculated by
q is charge;
ΔV is the potential difference;
Potassium ion has +1 charge, which means:
p = C
To determine work in joules, potential has to be in Volts, so:
Then, work is
To move a potassium ion from the exterior to the interior of the cell, it is required J of energy.
(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².
(b) The time when the velocity of the turtle is zero is 3.2 s.
(c) The time taken for the turtle to return to its starting point is 6.4 s.
(d) The time taken for the turtle to travel 30 cm is 0.08 s.
<h3>Initial velocity of the turtle</h3>The initial velocity of the turtle is calculated as follows;
The initial acceleration of the turtle is calculated as follows;
x(t) = 5 + 4t - 0.625t²
x(0) = 5 cm
<h3>Time when the velocity becomes zero</h3>v(t) = 4 - 1.25t
0 = 4 - 1.25t
1.25t = 4
t = 4/1.25
t = 3.2 s
<h3>Time taken to return to starting point</h3>The total distance traveled is calculated as follows
v² = u² + 2ad
0 = (4)² + 2(-1.25)d
0 = 16 - 2.5d
2.5d = 16
d = 16/2.5
d = 6.4 m
Time to travel the given distance;
d = ut + ¹/₂at²
6.4 = (4)t + ¹/₂(-1.25)t²
6.4 = 4t - 0.625t²
0.625t² - 4t + 6.4 = 0
solve the quadratic equation using formula method;
t = 3.2 s
The time travel the distance two times, = 2 x 3.2 s = 6.4 s
<h3>Time taken for the turtle to travel 30 cm</h3>d = ut + ¹/₂at²
0.3 = (4)t + ¹/₂(-1.25)t²
0.3 = 4t - 0.625t²
0.625t² - 4t + 0.3 = 0
solve the quadratic equation using formula method;
t = 0.08 s
Learn more about velocity here: brainly.com/question/6504879