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elena-14-01-66 [18.8K]
3 years ago
10

¿Por qué si cargas a uno de tus compañeros por cierto tiempo no estás realizando un trabajo mecánico?

Physics
1 answer:
VladimirAG [237]3 years ago
4 0

Answer:

I will answer this in English, we can translate it to:

Why if you charge a mate by an amount of time you are not doing work?

This happens because work is defined as the displacement done by a force:

W = d*F

where W is work, d is the distance, and F is the force.

This means that the amount of time that you are charging your mate does not affect the mechanical work, the only time that you are doing work is when you are lifting him.

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Acceleration is how much the velocity changes within a period of time so,

Acceleration= is the change in velocity divided by change in time

your units will be m/s squared
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Electromagnetic waves travel through space at a speed of _____.
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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

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6 0
1 year ago
Define compound and list two examples.
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3 years ago
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A parallel-plate capacitor in air has a plate separation of 1.76 cm and a plate area of
Monica [59]

Answer:

Explanation:

Plate separation, d = 1.76 cm = 0.0176 m

Area of plates, A = 25 cm^2 = 0.0025 m^2

V = 255 V

(a) Capacitance of capacitor

C = \frac{\epsilon _0A}{d}

C = \frac{8.854\times 10^{-12}\times 0.0025}{0.0176}

C = 1.258 x 10^-12 F

charge is same before and after immersion as the battery is disconnected

q = C V

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(b)

Capacitance before, C = 1.258 x 10^-12 C

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Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V

(c) initial energy,

U = \frac{q^{2}}{2C}

U = \frac{(3.2\times 10^{-10})^{2}}{2\times 1.258\times 10^{-12 }}=4.07\times 10^{-8}J

Final energy

U' = \frac{q^{2}}{2C'}

U' = \frac{(3.2\times 10^{-10})^{2}}{2\times 100.64\times 10^{-12}}=5.08\times 10^{-10}J

6 0
3 years ago
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