Answer:
<h3>1/16</h3>
Explanation:
According to the coulombs law, the force existing vetween the ions is expressed as;
F = kQq/r² .... 1
Q and q are the ions
r is the distance between the ions
If the distance between the ion is quadrupled, then;
F2 = kQq/(4r)²
F2 = kQq/16r² ... 2
Divide equation 2 by 1;
F2/F = kQq/16r² ÷ kQq/r²
F2/F = kQq/16r² × r²/kQq
F2/F = 1/16
F2 = 1/16 F
Therefore the coulombic force between two ions is reduced to<u> 1/16 </u>of its original strength when the distance between them is quadrupled.
Answer:
A) x4
Explanation:
Magnification is equal to image size divided by the actual size, or M = I/A.
The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:
28.8 cm/7.2 cm = 4
Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
Explanation:
6. Converge or come together
7. convex
Answer:
The answer to the question is 7200
