The magnitude of the work done by the electric field of the membrane is <u>W = 1.28 × 10⁻²⁰ Joules</u>.
We start with the necessity to take into account a value for the voltage present there in order to solve this problem by first considering that the membranes have two layers, one internal and one external, each responsible for producing a potential difference between the two levels.
As a result, in order to find a solution, it is necessary to take into account the potential difference between the two surfaces. In this instance, we'll assume a particular value for the load, but the recipient is free to substitute a different value if they prefer.
The product of the potential difference and the charge is used to define the work that an electric field performs. The charge of the potassium ion will be equal to that of its electron, so,
q = 1.6 × 10⁻¹⁹ Coulombs
Then the Work would be:
W = Vq
Here,
v = Potential difference
q = Charge
The 80mV potential difference we will have is quantified as follows:
W = (80mV (1V/1000mV))( 1.6 × 10⁻¹⁹ C)
W = 1.28 × 10⁻²⁰ Joules is the amount of work that the membrane's electric field has produced.
Find more on work done at : brainly.com/question/25573309
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Copied but correct answer.
Answer: C (She hasn't fastened the slide to the stage
Explanation:
In Microscopy even if the slide is not yet in focus,she will be able to see part of the slide which may appear as transparent image.Therefore this is a wrong answer.
If the microscope is not plugged it light can not come up.
If there are no specimen on the slide,the slide will still as transparent image at the focus.
No doubt Stella forgot to fasten the slide to the stage.,therefore the specimen on the slide was not in focus of the objective lens,rather she was focusing on the stage only,hence the visible white light.
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