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Pepsi [2]
3 years ago
8

Two small spheres spaced 20.0 cm apart have equal charge. how many excess electrons must be present on each sphere if the magnit

ude of the force of repulsion between them is 4.57 * 10-21 n?
Physics
1 answer:
uysha [10]3 years ago
8 0
<span>890 Coulomb's law is F = k * q1 * q2 / r^2 where F = force k = Coulomb's constant 8.99x10^9 N m^2/C^2 q1,q2 = signed charges r = distance between charges Since the charges are the same, let's simplify the equation, solve for q, then substitute the known values and calculate. F = k * q1 * q2 / r^2 F = k * q^2 / r^2 F*r^2 = k * q^2 F*r^2/k = q^2 sqrt(F*r^2/k) = q sqrt(4.57x10^-21 N * (0.2m)^2 / (8.99x10^9 N m^2/C^2)) = q sqrt((4.57x10^-21 N * 0.04m^2) / (8.99x10^9 N m^2/C^2)) = q sqrt((1.828x10^-22 N*m^2) / (8.99x10^9 N m^2/C^2)) = q sqrt(2.03337x10^-32 C^2) = q 1.42596x10^-16 C = q So each sphere has to have an excess of 1.42596x10^-16 Coulombs of electrons. A coulomb is 6.24150934x10^18 electrons, so let's do the multiplication: 1.42596x10^-16 * 6.24150934x10^18 = 8.9001426584664x10^2 = 890 So each sphere has an extra 890 electrons.</span>
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Answer:

4.79 J

Explanation:

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An object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m
Svetllana [295]

The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

<h3>Work done by the force experienced  by the object</h3>

The magnitude of the work done by force experience by the object is calculated as follows;

W = f.d

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The work done by the force is determined from the dot product of the force and the displacement of the object.

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W = (2abi + 3bj).(ai + bj)

W = (2a²b + 3b²)J

Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

The complete question is below:

The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.

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2 years ago
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ololo11 [35]

Answer:

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Explanation:

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Using the simple relationship

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