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Alex787 [66]
3 years ago
12

Which best describes how geothermal energy is used to make electricity?. . A.. Rocks are split to release energy that drives a s

team turbine attached to an electrical generator.. . B.. Heat energy from below the ground converts water to steam to drive a steam turbine attached to an electrical generator.. . C.. Heat energy from below the ground drives liquid water though a turbine attached to an electrical generator.. . D.. Heat energy from below ground heats and presuurizes air that drives a turbine attached to an electrical generator..
Physics
1 answer:
In-s [12.5K]3 years ago
6 0
Answer is b that is  Heat energy from below the ground converts water to steam to drive a steam turbine attached to an electrical generator.. . 
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A common cylindrical copper wire used in a lab is 841 m long. Find the radius (in mm) of a wire necessary to have 0.5 Ohms of re
slava [35]
The relationship between the resistance R of a wire and its resistivity \rho is given by
R=  \frac{\rho L}{A}
where L is the length of the wire and A is its cross sectional area.

In the problem, we have R=0.5 \Omega, \rho = 1.68 \cdot 10^{-8} \Omega m and L=841 m. So we can solve the find the area A:
A= \frac{\rho L}{R}=2.83 \cdot 10^{-5} m^2

For a cylindrical wire, the cross sectional area is given by
A= \pi r^2
where r is the radius. We know the value of the area A, so now we can find the radius of the wire:
r= \sqrt{ \frac{A}{\pi} }= \sqrt{ \frac{2.83 \cdot 10^{-5}m^2}{\pi} }=0.003 m=3.0 mm
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3 years ago
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

#SPJ1

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2 years ago
Which of the following is true about semi-conductors: A. Exposing a crystal of a semiconductor to heat or light starts displacin
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Answer:

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