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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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Answer: Technician B is right.

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
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According to the free-body diagram of the system, we have:

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So, we can solve for T from (1):

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The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

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mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C

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