An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 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 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

1.been both -ve charged or both +be charged particles

2. 3.52mC

Explanation:

For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.

Now the Force sustain by the extended string is

F = Ke;

Where K is the force constant of the string, 320 N/m

e is the extension,0.033 m

F = 320 × 0.033 =10.56N

2.But according to columns law of charge;

F = kQ1 Q2

But Q1=Q2{ since the charge are of the same magnitude}.

Hence F = KQ^2

Where K is columns constant =9×10^9F/m

Hence Q=√F/K

Q= √10.56/9×10^9

=3.52×10^-3C

= 3.52mC

emily ruth's annual premium is $1058.50

step-by-step explanation:

emily ruth is a female of 27 years age purchasing 20 year endowment insurance with face value $25000.

from the given table,

for females, the age of a male 3 years younger should be used.

so, we have to check for 24 year male for 20 year endowment.

it is 42.34 for every $1000 face value.

for a face value of $25000,

it is 25 * 42.34 = $1058.50

∴ emily ruth's annual premium is $1058.50

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