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madam [21]
4 years ago
5

Anita and Nick are playing tug-of-war near a mud puddle. They are each holding on to an end of a taut rope that has a knot exact

ly in the middle. Anita's position is 6.2 meters east of the center of the puddle and Nick's position is 3.0 meters west of the center of the puddle. What is the location of the knot relative to the center of the puddle? Treat east as positive and west as negative.

Physics
1 answer:
Vadim26 [7]4 years ago
6 0

Answer:

1.6 meters east

Explanation:

Note: Image attached.

As we can see on the data, Anita is further from the puddle than Nick. Additionally, the distances relatives to the center of the puddle allow us to now the lenght of the rope which is the sum of these two, 9.2 meters. We also know that the knot is at the middle of the rope, so it means it is at 4.6 meters of  both Anita and Nick. Now what is next? Well, we now have everything we know. We have the distance between Anita and the puddle and we also know the distance between Anita and the knot (4.6 meters). We just need to subtract the distance to the knot from the distance to the puddle which is equal to 1,6 meters east direction.

If you guys have any questions, just let me know.

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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
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  • (a) Q = 1.70\times 10^{-2}\;\text{C};
  • (b) V_\text{final} = 5.31\times 10^{2}\;\text{V};
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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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