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3 years ago

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

Physics

1 answer:

77julia77 [94]3 years ago

7 0

Answer:

The separation distance between the two charges must be 82704.2925 m

Explanation:

Given:

Two negative charges that are both q = -3.8 C

Force of 19 N

Question: How far apart are the two charges, s = ?

First, you need to get the electrostatic force of this two negative charges:

$F=\frac{kq}{s^{2} } \\s=\sqrt{\frac{kq}{F} }$

Here

k = electric constant of the medium = 9x10⁹N m²/C²

Substituting values:

$s=\sqrt{\frac{9x10^{9}*(-3.8)^{2} }{19} } =82704.2925m$

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Three dogs (Spot, Fido, and Steinberg) are pulling on a chew toy. The chew toy is experiencing no acceleration. Spot is pulling

quester [9]

Answer:

The magnitude and direction of the force applied by Steinberg are approximately 15.192 newtons and 126.704º.

Explanation:

The chew toy is at equilibrium and experimenting three forces from three distinct dogs. The Free Body Diagram depicting the system is attached below. By Newton's Laws we construct the following equations of equilibrium: (Sp is for Spot, F is for Fido and St is for Steinberg) All forces and angles are measured in newtons and sexagesimal degrees, respectively:

$\Sigma F_{x} = F_{F}\cdot \cos \theta_{F} + F_{St,x} = 0$ (1)

$\Sigma F_{y} = F_{F}\cdot \sin \theta_{F}-F_{Sp}+F_{St,y} = 0$ (2)

If we know that $F_{F} = 20\,N$ , $F_{Sp} = 30\,N$ and $\theta_{F} = 63^{\circ}$ , then the components of the force done by Steinberg on the chewing toy is:

$F_{St,x} = -F_{F}\cdot \cos \theta_{F}$

$F_{St,x} = -(20\,N)\cdot \cos 63^{\circ}$

$F_{St, x} = -9.080\,N$

$F_{St,y} = F_{Sp}-F_{F}\cdot \sin \theta_{F}$

$F_{St,y} = 30\,N-(20\,N)\cdot \sin 63^{\circ}$

$F_{St, y} = 12.180\,N$

The magnitud of the force is determined by Pythagorean Theorem:

$F_{St} = \sqrt{F_{St,x}^{2}+F_{St,y}^{2}}$

$F_{St} =\sqrt{(-9.080\,N)^{2}+(12.180\,N)^{2}}$

$F_{St} \approx 15.192\,N$

Since the direction of this force is in the 3rd Quadrant on Cartesian plane, we determine the direction of the force with respect to the eastern semiaxis:

$\theta_{St} = 180^{\circ} + \tan^{-1} \left(\frac{F_{St,y}}{F_{St,x}}\right)$