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allsm [11]
3 years ago
12

Consider an ideal spring, with spring constant k, which is oriented along an x-axis. one end of the spring is fixed, and the fre

e end is located at x = 0 when the spring is neither stretched nor compressed. an object having a mass m is attached to the free end of the spring. consider an action in which the free end of the spring is moved from an initial position xi to a final position xf. the work done by the spring (and on the object attached to the free end) during this action is given by
Physics
1 answer:
Blizzard [7]3 years ago
5 0
The spring is neither stretched nor compressed. an object having a mass m is attached to the free end of the spring. consider an action
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A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to
liberstina [14]
  • (a) Q = 1.70\times 10^{-2}\;\text{C};
  • (b) V_\text{final} = 5.31\times 10^{2}\;\text{V};
  • (c) E_\text{final} = 4.52\;\text{J};
  • (d) \Delta E = 2.82\;\text{J}.

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}.

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial};

(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial};

\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}.

(c)

\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}.

\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}.

(d)

Initial energy of the system, which is the same as the initial energy in the 20.0\;\mu\text{F} capacitor:

\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}.

Change in energy:

\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}.

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