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How much heat<span> is </span>required<span> to </span>heat 0.1 g<span> of </span>∆hvap<span> =</span>2260 j/g ∆h<span> =</span>340j/g fus iceat−30 ctosteamat 100c?use<span> the </span>approximate values<span>?</span>
- The equivalent capacitance between point a and b is 5.87μf.
- The charge at 20μf is 93.92 μC.
- The charge at 6μf is 67.8 μC.
- The charge at C and 3μf is 26.12 μC.
<h3>
Sum of capacitance of C and 3μF</h3>
The sum of the capacitance is calculated as follows;
1/Ct = 1/C + 1/3
1/Ct = 1/10μf + 1/3μf
1/Ct = (3μf + 10μf)/30μf²
1/Ct = 13μf/30μf²
Ct = 30μf²/13μf
Ct = 2.31μf
<h3>Total capacitance in parallel arrangement</h3>
The total capacitance in parallel arrangement is calculated as follows;
Ct = 2.31μf + 6μf = 8.31μf
<h3>Equivalent capacitance between point a and b</h3>
1/Ct = 1/8.31μf + 1/20μf
1/Ct = 0.1703
Ct = 5.87μf
<h3>Charge flowing in each capacitor</h3>
Maximum voltage is delivered in 20μf, q = CV
<u>charge for 20μf:</u>
q = (5.87 x 16)μC
q = 93.92 μC
<h3>Equivalent capacitance for C, 3μf and 6μf</h3>
Ct = 2.31μf + 6μf = 8.31uf
<u>charge for 6μf:</u>
q = (6/8.31) x 93.92μC
q = 67.8μC
<h3>Total charge for C and 3μf</h3>
q = 93.92μC - 67.8μC = 26.12 μC
charge for C = charge 3μf = 26.12 μC
Learn more about capacitor here: brainly.com/question/14883923
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As per given conditions there are two directions along which forces are acting
1. Net force along left direction is given as

2. Net force towards right direction is given as

now since the two forces here in opposite direction so here we will have net force given as



so here net forces must be 440 N towards right