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MA_775_DIABLO [31]
3 years ago
13

It turns out that most of the electricity we use in Ohio comes from burning coal. Coal yields about 30 Gigajoules (GJ = 109 J) o

f electricity per metric ton (1000 kg). Assuming that the coal plant is 30% efficient (i.e. the plant captures 30% of the energy converted by burning coal), how much total coal has to be burned to keep that 100 W light bulb lighted for the week?
Physics
1 answer:
rodikova [14]3 years ago
5 0

Answer:

0.6048 kg of coal

Explanation:

Mass energy input = mass energy output/efficiency

Mass energy output = 30 GJ/ton= 3×10^10 J/1000 kg = 3×10^7 J/kg

Efficiency = 30% = 0.3

Mass energy input = 3×10^7/0.3 = 1×10^8 J

Mass of coal to be burned = power×time/mass energy input

Power = 100 W

Time = 1 week = 604,800 s

Mass of coal to be burned = 100×604,800/1×10^8 = 0.6048 kg

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio
LUCKY_DIMON [66]

Answer:

29274.93096 m/s

2.73966\times 10^{33}\ J

-5.39323\times 10^{33}\ J

2.56249\times 10^{33}\ J

-5.21594\times 10^{33}

Explanation:

r_p = Distance at perihelion = 1.471\times 10^{11}\ m

r_a = Distance at aphelion = 1.521\times 10^{11}\ m

v_p = Velocity at perihelion = 3.027\times 10^{4}\ m/s

v_a = Velocity at aphelion

m = Mass of the Earth =  5.98 × 10²⁴ kg

M = Mass of Sun = 1.9889\times 10^{30}\ kg

Here, the angular momentum is conserved

L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J

Kinetic energy at perihelion is 2.73966\times 10^{33}\ J

Potential energy is given by

P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times  10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}

Potential energy at perihelion is -5.39323\times 10^{33}\ J

K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J

Kinetic energy at aphelion is 2.56249\times 10^{33}\ J

Potential energy is given by

P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}

Potential energy at aphelion is -5.21594\times 10^{33}\ J

6 0
3 years ago
Tres personas, A, B, C, jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección ho
Yuri [45]

Answer:

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

Explanation:

Si la caja debe hallarse en equilibrio físico, entonces se debe satisfacer la siguiente ecuación:

F_{A} + F_{B} + F_{C} = 0 (1)

Si sabemos que F_{A} = -3 y F_{B} = 5, entonces el valor de la fuerza que debe ejercer la persona C debe ser:

F_{C} = -F_{A}-F_{B}

F_{C} = -(-3)-5

F_{C} = -2

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

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3 years ago
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Explanation: Evaporation happens when a liquid substance becomes a gas. When water is heated, it evaporates. The molecules move and vibrate so quickly that they escape into the atmosphere as molecules of water vapor (Dunn,1993).
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timofeeve [1]

Answer:

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Explanation:

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