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cluponka [151]
3 years ago
9

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.
Physics
1 answer:
lakkis [162]3 years ago
8 0

Answer:

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

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a) Increase

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a)

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In this second part, the battery voltage is doubled.

According to the equation written in part a),

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We see that this equation does not depend at all on the voltage of the battery.

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We know that for two resistors in series, the total resistance of the circuit is given by the sum of the individual resistances:

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Since each resistance is a positive value, this means that as we add new resistors, the total resistance of the circuit increases.

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The time taken for the capacitor to reach 90% of its final charge is still

t=2.30 RC

As we can see, this time is directly proportional to the resistance of the circuit, R: therefore, if we add a resistor in series, the resistance of the circuit will increase, and therefore this time will increase as well.

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