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Tcecarenko [31]
3 years ago
10

Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran

Physics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

Datos:

q1 = -50 μC = 50*10^{-6}

q2 = +30 μC = 30*10^{-6}

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = \sqrt\frac{k0 * q1 *q2}{F} = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10} = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em>\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}<em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50} = 0,520m

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Answer:

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(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

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ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

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When positively charged particles were radiated onto a gold atom, a few of the particles bounced back. Which of the following ca
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Calculate the workdone to stretch an elastic string by 40cm if a force of 10 newton produces an extension of 4cm in it
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Answer:

Workdone = 20 Joules

Explanation:

Given the following data;

Force = 10N

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First of all, we would find the spring constant using the formula;

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