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Kamila [148]
4 years ago
15

Explain why there is no acceleration horizontally on an object in Projectile Motion. A. There is no acceleration horizontally be

cause velocity is constant in the x- direction. B. There is no acceleration horizontally because velocity increases in the x- direction. C. There is no acceleration horizontally because velocity decreases in the x-direction . D. There is no acceleration vertically because velocity is constant in the y -direction .
Physics
1 answer:
Jobisdone [24]4 years ago
4 0

Answer:

Option  A is correct.

Explanation:

The force of gravity does not affect the horizontal component in projectile motion; a projectile maintains a constant horizontal velocity. So there is no acceleration ( a=0 ) in horizontal direction as velocity remains constant.

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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
fenix001 [56]

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

V=\frac{Q}{C}=\frac{0.0071298 C}{9.42\times 10^{-6} F}=756.88 V

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What is potential difference?<br>Define Ohm's law.​
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Answer:

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