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natali 33 [55]
3 years ago
14

Is oxygen a reactant or product

Physics
2 answers:
jok3333 [9.3K]3 years ago
5 0

Answer:

reactant

Explanation:

dezoksy [38]3 years ago
3 0

Answer:

Methane and oxygen (oxygen is a diatomic — two-atom — element) are the reactants, while carbon dioxide and water are the products

Explanation:

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Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de
ahrayia [7]

Answer:

a) El caudal de salida del chorro es 1.213\times 10^{-3}\,\frac{m^{3}}{s}.

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

\Delta z = \frac{v_{out}^{2}}{2\cdot g}

Donde:

\Delta z - Diferencia de altura, medida en metros.

g - Constante gravitacional, medida en metros por segundo al cuadrado.

v_{out} - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

v_{out} = \sqrt{2\cdot g \cdot \Delta z}

Si g = 9.807\,\frac{m}{s^{2}} y \Delta z = 0.3\,m, entonces la rapidez de salida del chorro es:

v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}

v_{out} \approx 2.426\,\frac{m}{s}

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

\dot V_{out} = v_{out}\cdot A_{t}

Donde:

v_{out} - Rapidez de salida del chorro, medida en metros por segundo.

A_{t} - Área transversal del orificio, medido en metros cuadrados.

\dot V_{out} - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que v_{out} = 2.426\,\frac{m}{s} y A_{t} = 5\,cm^{2}, el caudal de salida del chorro es:

\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}}  \right)

\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}

El caudal de salida del chorro es 1.213\times 10^{-3}\,\frac{m^{3}}{s}.

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A 40.0-μFcapacitor is connected across a 60.0 Hz generator. An inductor is then connected in parallel with the capacitor. What i
amm1812

Answer:

The value of the inductance is 175.9 mH.

Explanation:

Given that,

Capacitor C= 40.0\ \muF

Frequency = 60.0 Hz

The inductor and capacitor is connected in parallel, the voltage across each of these elements is the same.

We have,

V_{L}=V_{C}

Using ohm's law

I_{rms}\times X_{L}=I_{rms}\times X_{C}

X_{L}=X_{C}

2\pi f L=\dfrac{1}{2\pi f C}

L=\dfrac{1}{4\pi^2\times f^2\times C}

Put the value into the formula

L=\dfrac{1}{4\times\pi^2\times60.0^2\times40\times10^{-6}}

L=0.1759\ H

L=175.9\times10^{-3}\ H

L=175.9\ mH

Hence, The value of the inductance is 175.9 mH.

5 0
4 years ago
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