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3 years ago

Please calculate the current for the circuit below

Engineering

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

The answer is " $I = 0.0085106383 \ A$ "

Explanation:

Given:

$R= 470 \ \Omega \\\\V= 4 \ v$

Formula:

$\to V=IR\\\\\to I = \frac{V}{R}\\\\$

$= \frac{4}{470}\\\\ = 0.0085106383 \ A$

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storchak [24]

Answer: drugs and rushing cars

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3 years ago

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve

Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature $T_1 = -16^0\ C$

Quality $x_1 = 0.2$

Outlet:

Temperature $T_2 = -16^0 C$

Quality $x_2 = 1$

The following data were obtained at saturation properties of R134a at the temperature of -16° C

$v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg$

$v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg$

$m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}$

$\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}$

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3 years ago

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Answer:

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Explanation:

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3 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 16.403

Dmitry [639]

Answer:

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Explanation:

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4 years ago

What two things must be included in your function definition?

mars1129 [50]

Explanation:

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