Answer:
oone side at a time would be hot and the other oone
Explanation:
One side will be hot and the other cold no in-between
(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
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Answer:
is high as 100 degrees c
Explanation:
due to high heat gas expands fast than normal
<span>Density is 3.4x10^18 kg/m^3
Dime weighs 1.5x10^12 pounds
The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so
4/3 pi 1.9x10^3
= 4/3 pi 6.859x10^3 m^3
= 2.873x10^10 m^3
Now divide the mass by the volume
9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3
Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3
Now to figure out how much the dime weighs, just multiply by the volume of the dime.
3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg
And to convert from kg to lbs, multiply by 2.20462, so
6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
