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shtirl [24]
3 years ago
10

Balance this chemical equation Fe(s) + 02(g) --- Fe0(s)

Physics
1 answer:
Softa [21]3 years ago
6 0
2Fe(s) + O2 -> 2FeO(s) 

<span>2 'Fe' atoms on both sides </span>
<span>2 'O' atoms on both sides</span>
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What are the requirement for a valid hypothesis ?
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3 0
3 years ago
You have a circuit with a 50 Ω , a 100 Ω , and a 150 Ω - resistor connected in series. (a) Rank the current through them from hi
olga2289 [7]

Explanation:

(a) A circuit has 50 Ω, a 100Ω and a 150 Ω resistor are connected in series. We know that in series combination, current through each resistor is same. So, current through a 50 Ω, a 100Ω and a 150 Ω resistor is same.

(b) Ohm's law of given by :

V = I R

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In the Bathburst 1000, racing cars take 6.25 hours to complete the 1000km race. Race is their average speed?
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160

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3 years ago
The cytoplasm is the watery fluid found within cells. The cytoplasm holds all of the organelles, except _______, in place within
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3 years ago
Un montañero de 65kg de masa ha ascendido a la cima del Everest, la montaña más alta del mundo de 8848m de altura sobre el nivel
andrew-mc [135]

Answer:

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

Explanation:

Por la definición de trabajo sabemos que el montañero debió contrarrestar trabajo causado por la gravedad terrestre. Si asumimos que el cambio de la altura es muy pequeño en comparación con el radio del planeta (6371 kilómetros vs. 0,5 kilómetros), entonces podemos considerar que la aceleración gravitacional es constante y la ecuación de trabajo (\Delta W), medido en joules, que reducida a:

\Delta W = m\cdot g\cdot \Delta z (1)

Donde:

m - Masa del montañero, medido en kilogramos.

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

\Delta z - Distancia vertical de ascenso del montañero, medida en metros.

Si tenemos que m = 65\,kg, g = 9,807\,\frac{m}{s^{2}} y \Delta z = 500\,m, entonces el trabajo realizado por el montañero para subir es:

\Delta W = (65\,kg)\cdot \left(9,807\,\frac{m}{s^{2}} \right)\cdot (500\,m)

\Delta W = 318727,5\,J

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

7 0
3 years ago
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