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Paha777 [63]
3 years ago
13

Part AIf the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negativ

e terminal of the battery is at zero potential.Part BIf the charge of the first capacitor (the one with capacitance C) is Q, then what are the charges of the second and third capacitors?Part CSuppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of the three individual capacitors calculated in the previous part, find the total charge Qtot for this equivalent capacitor.Part DUsing the value of Qtot, find the equivalent capacitance Ceq for this combination of capacitors.Express your answer in terms of C.

Physics
1 answer:
Nana76 [90]3 years ago
6 0

Answer:

Part a: The potential at point 3 and point 6 are V and 0 respectively.

Part b: The charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c: The net charge is  6CV.

Part d: The equivalent capacitance is 6C

Explanation:

As the question is not given here ,the complete question is found online and is attached herewith.

Part a:

V1 = V, V3 = V1 = V

and, V6 = V1-V = 0

The potential at point 3 and point 6 are V and 0 respectively

Part b

Q1 =CV = Q,

Q2 = 2C *V = 2Q

Q3 = 3 C*V = 3Q

So the charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c

Total charge of the system,

Q_net =Q1+Q2+Q3= (1+2+3) CV = 6 CV

So the net charge is  6CV.

Part d:

Equivalent Capacitance = Net charge / Voltage

Eq. C = 6CV/V = 6C

So the equivalent capacitance is 6C

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Answer:

The mass of the aluminum chunk is 258 g

Explanation:

Given;

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initial temperature of the copper cube, T_c_u = 85°C

initial temperature of the aluminum chunk T_A_l = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}

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M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg

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7 0
3 years ago
How much metabolic energy is required for a 68kg person to run at a speed of 15km/hr for 15min ?
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<span>What I have here is exactly the same problem, however, with the time changed to 19 mins:

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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
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v_2=-133.17m/s, the minus meaning west.

Explanation:

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After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

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