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lyudmila [28]
3 years ago
8

A liquid of density 1390 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the

flow speed is 9.63 m/s and the pipe diameter d 1 is 10.1 cm . At Location 2, the pipe diameter d 2 is 15.3 cm . At Location 1, the pipe is Δ y = 8.85 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ P between the fluid pressure at Location 2 and the fluid pressure at Location 1.
Physics
1 answer:
Aloiza [94]3 years ago
6 0

Answer:

The difference of power is

ΔP = 172.767 kPa

Explanation:

ρ = 1390 kg / m³

v = 9.63 m/s

d₁ = 10.1 cm , d₂ = 15.3 cm

Δz = 8.85 m

To find the difference ΔP between the fluid pressure at locations 2 and the fluid pressure at location 1

ΔP = ρ * g * Z + ¹/₂ * ρ * v² * ( 1 - (d₁ / d₂)⁴ )

ΔP = 1390 kg / m³ * 9.8 m/s² * 8.85 m + 0.5 * 1390 kg / m³ *(9.63 m /s)² * (1 - (0.101 m / 0.153 m )⁴ )

ΔP = 172.767 x 10 ³ Pa

ΔP = 172.767 kPa

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Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r
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Answer:

  • Current = 0.33 A

Explanation:

  • For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

\dashrightarrow \: \:  \sf V = 2 \times 5 = 10 V

Three resistor of 5\Omega, 10\Omega, 15\Omega are connected in Series, so the net resistance:

\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}

\dashrightarrow \: \:  \sf R = 5 + 10 + 15

{ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \:  \Omega}}}}

According to ohm's law:

\dashrightarrow  \sf\: \: V = IR

\dashrightarrow  \sf \: \: I = \dfrac{V}{R}

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 {\pmb{\sf{\Omega}}} we get:

\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}

{ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}

\thereforeThe electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

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A sample of helium gas has a volume of 900. milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the t
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<em>The new pressure = 5.64 atm.</em>

Explanation:

Using The general gas equation,

P₁V₁/T₁ = P₂V₂/T₂..................... Equation 1

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P₂ = P₁V₁T₂/V₂T₁ ...................... Equation 2

Where P₁ = initial pressure, V₁ = initial Volume, T₁ = Initial temperature, P₂ = final pressure, V₂ = final volume, T₂ = final Temperature.

<em>Given: P ₁= 2.5 atm, T₁ = 298 K, V₁= 900 milliliters, T ₂= 336 K, V₂ = 450 milliliters</em>

<em>Substituting these values into equation 2,</em>

<em>P₂ = (2.5×900×336)/(298×450)</em>

P₂ = 756000/134100

P₂ = 5.64 atm.

<em>Thus the new pressure = 5.64 atm.</em>

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