An unstrained horizontal spring has a length of 0.43 m and a spring constant of 238 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.013 m relative to its unstrained length. Determine the possible algebraic signs and the magnitude of the charges. (a) the possible algebraic signs

The charges on the spring are 1.23E-5 C and they have the same sign

Explanation:

Given

Coulomb laws states that the force exerted by a charge q on another charge Q at a distance r is given by

F = kqQ/r²

Where k = 8.99 * 10^9 Nm²/C²

r = 0.43 + 0.013

r = 0.443m

The force on the spring is calculated as;.

F = kx where x is the stretch length of the spring and k is the spring constant

The force acting on the spring = 238 * 0.013

F = 3.094N

By comparison;

F = kqQ/r² becomes

3.094 = F = kqQ/r²

kqQ/r² = 3.094 (Considering that a = Q)

kq²/r² = 3.094

8.99 * 10^9 * q²/0.443 = 3.094 -- make q² the subject of formula

q² = 3.094 * 0.443/8.99*10^9

q² = 1.524629588431E−10

q = √1.524629588431E−10

q = 0.000012347589191545C

q = 1.23E-5 C

The charges on the spring are 1.23E-5 C and they have the same sign

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the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.