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Sav [38]
3 years ago
8

Characteristics sa Ancient art

Physics
1 answer:
yuradex [85]3 years ago
7 0

Answer:

Ancient art refers to the many types of art produced by the advanced cultures of ancient societies with some form of writing, such as those of ancient China, India, Mesopotamia, Persia, Palestine, Egypt, Greece, and Rome. The art of pre-literate societies is normally referred to as Prehistoric art and is not covered here. Although some Pre-Columbian cultures developed writing during the centuries before the arrival of Europeans, on grounds of dating these are covered at Pre-Columbian art, and articles such as Maya art and Aztec art. Olmec art is mentioned below.

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A 2.81 μF capacitor is charged to 1220 V and a 6.61 μF capacitor is charged to 560 V. These capacitors are then disconnected fro
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Answer:

756.88 Volts will be the potential difference across each capacitor.

Explanation:

Q=C\times V

Q = Charge on capacitor

C = Capacitance

V = Voltage across capacitor

Capacitance of first capacitor = C_1=2.81 \mu F=2.81\times 10^{-6} F

Charge of first capacitor = Q_1

Voltage across first capacitor = V_1=1220 V

Q_1=C_1V_1

Q_1=2.81\times 10^{-6} F\times 1220 V=0.0034282 C

Capacitance of first capacitor = C_2=6.61\mu F=6.61\times 10^{-6} F

Charge of second capacitor = Q_2

Voltage across first capacitor = V_2=560 V

Q_2=C_2V_2

Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C

Total capacitance in parallel combination:

C_p=C_1+C_2=2.81\times 10^{-6} F+6.61\times 10^{-6} F=9.42\times 10^{-6} F

Potential across both capacitors = V

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