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3 years ago

5. Two negative charges that are both - 3.8 C push each other apart with a force of 19.0 N. How far apart

Physics

1 answer:

V125BC [204]3 years ago

4 0

Answer:

82700 m

Explanation:

Applying Coulomb's law of electrostatic

F = kq²/r².......................... Equation 1

Note: Both charges are the same

Where F = force of attraction, q = negative charge, r = distance between the charges, k = coulomb's constant

make r the subject of the equation

r = √(kq²/F).............. Equation 1

Given: F = 19 N, q = 3.8 C, k = 9×10⁹ Nm²/C²

Substitute these values into equation 2

r = √(3.8²×9×10⁹/19)

r = √(68.4×10⁸)

r = 8.27×10⁴ m

r = 82700 m

Hence the two charges are 82700 m apart

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The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

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A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

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$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

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$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

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$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

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so, t=0.

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So, the time needed is 3 seconds.

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