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Naddik [55]
3 years ago
15

A stream moving with a speed of 7.1 m/s reaches a point where the cross-sectional area of the stream decreases to one half of th

e original area. What is the speed of the water in this narrowed portion of the stream?
Physics
1 answer:
lana66690 [7]3 years ago
8 0

Answer:

14.2 m/s

Explanation:

Given data:

Speed of the stream, v₁ = 7.1 m/s

let the cross section area at initial point be A₁

now area at the second point, A₂ = (1/2)A₁ = 0.5A₁

now, from the continuity equation, we have

A₁v₁ = A₂v₂

where, v₂ is the velocity at the narrowed portion

thus, on substituting the values, we get

A₁ × 7.1 = 0.5A₁ × v₂

or

v₂ = 14.2 m/s

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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:

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Q_1=C_1V_1

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

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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.

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Total capacitance in parallel combination:

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756.88 Volts will be the potential difference across each capacitor.

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