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vovangra [49]
3 years ago
8

A pump and its horizontal intake pipe are located 82 m beneath the surface of a large reservoir. the speed of the water in the i

ntake pipe causes the pressure there to decrease, in accord with bernoulli's principle. assuming nonviscous flow, what is the maximum speed with which water can flow through the intake pipe? (assume atmospheric pressure is 1.01 105 pa.)
Physics
1 answer:
uysha [10]3 years ago
4 0
<span>Ans : Bernoulli's principle states for incompressible non-viscous flow that p/Ď + gâ™h + (1/2)â™v² = constant Evaluate the equation along a stream line from liquid surface of the reservoir (1) to the inlet of the pipe pâ‚/Ď + gâ™hâ‚ + (1/2)â™v₲ = pâ‚‚/Ď + gâ™hâ‚‚ + (1/2)â™v₂² => vâ‚‚ = âš[ 2â™(pâ‚-pâ‚‚)/Ď + 2â™gâ™(hâ‚-hâ‚‚) + v₲ ] lets make some assumptions: - the pressure at the liquid surface is equal to the atmospheric pressure pâ‚ = 1atm = 101325Pa - the velocity of the liquid at the surface (that is the speed at which the liquid level in reservoir decreases) is quite small, so it may be ignored: v₠≠0 So vâ‚‚ = âš[ 2â™(pâ‚-pâ‚‚)/Ď + 2â™gâ™(hâ‚-hâ‚‚) ] The height difference is fixed. So the only variable remaining is the pressure in the pipe. As higher it is as lower the velocity in the pipe is. So you get the maximum velocity for the minimum pressure. Since pressure cannot drop below zero this is pâ‚‚ = 0 Therefore vâ‚‚max = âš[ pâ‚/Ď + gâ™(hâ‚-hâ‚‚) ] = âš[ 2â™101325Pa/1000kg/mÂł + 2â™9.81m/s²â™12m ] = 20.93m/s</span>
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Answer:

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Explanation:

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3 years ago
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A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b)
Alexeev081 [22]

Answer:

a) 0.01232 A

b) 0.00075 s = 0.75 ms

c) 0.0045323 A = 4.532 mA

d) 3.894 V

Explanation:

R = 500 Ω

V = 6.16 V

C = 1.50 μF

Let Vs be the voltage of the emf source

Let Vc be the voltage across the capacitor at any time

a) Current flows as a result of potential difference between two points. So, the current flows according to difference in voltage between the emf source and the capacitor.

At time t = 0,

There is no voltage on the capacitor; Vc = 0 V

Current in the circuit is given by

I = (Vs - Vc)/R

I = (6.16 - 0)/500

I = 0.01232 A

b) Time constant for an RC circuit is given by τ

τ = RC = (500) (1.5 × 10⁻⁶) = 0.00075 s

c) The current decay in an RC circuit (called decay because the current in the circuit starts to fall as the capacitor's voltage rises as the capacitor charges) is given by

I = I₀ e⁻ᵏᵗ

where k = (1/τ)

I₀ = Current in the circuit at t = 0 s; I₀ = 0.01232 A

At t = τ = 0.00075 s, kt = (τ/τ) = 1

I = 0.01232 e⁻¹ = 0.0045323 A = 4.532 mA

d) The voltage for a charging capacitor is given by

Vc = Vs (1 - e⁻ᵏᵗ)

where k = (1/τ)

At t = τ = 0.00075 s, Vc = ?, Vs = 6.16 V, kt = 1

Vc = 6.16 (1 - e⁻¹) = 6.16 (0.6321)

Vc = 3.894 V

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3 years ago
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Answer:

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Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\

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(b)

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3 years ago
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Answer:

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The work is negative (W-) if the force has the opposite direction of the movement of the object.

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W = 150 J

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3 years ago
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