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3 years ago

What is the electric force acting between two charges of -0.0050 C and

Physics

1 answer:

love history [14]3 years ago

6 0

The electric force is $-3.6\cdot 10^8 N$ (attractive)

Explanation:

The magnitude of the electric force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have the following:

$q_1 = -0.0050 C$ (charge 1)

$q_2 = +0.0050 C$ (charge 2)

r = 0.025 m (distance)

Substituting, we find the electric force between the two charges:

$F=(9\cdot 10^9) \frac{(-0.0050)(0.0050)}{(0.025)^2}=-3.6\cdot 10^8 N$

And the negative sign means the force is attractive, since the two charges have opposite sign.

Learn more about electric force:

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brainly.com/question/4273177

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Read 2 more answers

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$