Given: Each block has masses m1 = m2 = m3 = m and the coefficient of kinetic friction is µ. The magnitude of F equals twice the
total frictional force. Apply a horizontal force F in pushing an array of three identical blocks in the horizontal plane (see sketch). m1 m2 m3 F µ Find the force F23 with which the second block is pushing the third block.
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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.
The Electric field is zero at a distance 2.492 cm from the origin.
Let A be point where the charge C is placed which is the origin.
Let B be the point where the charge C is placed. Given that B is at a distance 1 cm from the origin.
Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.
i.e., at distance 'x' from B.
Using Coulomb's law, ,
=
k is the Coulomb's law constant.
On substituting the values into the above equation, we get,
Taking square roots on both sides and simplifying and solving for x, we get,
1.67x = 1+x
Therefore, x = 1.492 cm
Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.
Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926
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