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2 years ago

What is the potential energy of two charges of +4.6 μC and +1.0 μC that are separated by a distance of 10.0 cm?

Physics

1 answer:

Artist 52 [7]2 years ago

3 0

Answer:

U = 0.413 J

Explanation:

the potential energy between two charges q1 and q2 is given by the following formula:

$U=k\frac{q_1q_2}{r}$ (1)

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q1: first charge = 4.6 μC = 4.6*10^-6 C

q2: second charge = 1.0 μC*10^-6 C

r: distance between charges = 10.0 cm = 0.10 m

You replace the values of all variables in the equation (1):

$U=(8.98*10^9Nm^2/C^2)\frac{(4.6*10^{-6}C)(1.0*10^{-6}C)}{0.10m}=0.413\ J$

Hence, the energy between charges is 0.413 J

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3 years ago

In an exciting game, a baseball player manages to safely slide into second base. The mass of the baseball player is 88.1 kg and

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Answer:

$f=431.69\ N$

Explanation:

Given:

mass of the player, $m=88.1\ kg$

coefficient of kinetic friction between the ground and the player, $\mu_k=0.5$

Since the normal reaction force on the player due to its mass will be equal to the weight of the player.

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2 years ago

A charge of 3.20 μC is held fixed at the origin. A second charge of 3.20 μC is released from rest at the position (1.25 m, 0.570

siniylev [52]

Answer:

a) $v = 7.137 m/s$ and b) $r = 1.832\ m$

Explanation:

We have to analyze this problem from the point of view of energy conservation. In this case there are two kind of energy, electric potential energy and kinetic energy. First, at $t = t_1$ there isn't relative movement between the two charges, so kinetic energy is zero and the total energy ( $E_T = E_p + E_k$ ) is just potential.

$E_T = E_p = \frac{k q_1 q_2}{r}$

where $k$ is the Coulomb constant, $q_1$ and $q_2$ are the two interacting charges, and $r$ is the distance between them.

Considering the fixed charge at (x1,y1) = (0,0) and the second one at (x2, y2) = (1.25, 0.57), the initial distance is

$r = \sqrt{(x_2-x_1)^2 + (y_2- y_1)^2} = \sqrt{(1.25)^2 + (0.57)^2} = 1.374\ m$ , then if

$k =8.987\times 10^9 N m^2/C^2$ and $q_1 = q_2 =3.2\times 10^{-6} C$ ,

$E_T = E_p = \frac{k q_1 q_2}{r} = \frac{8.987\times 10^9 * 3.2\times 10^{-6}*3.2\times 10^{-6}}{1.374} = 6.698 \times 10^{-2} Nm$ .

Now, at $t = t_2$ , $r \rightarrow \infty$ and $E_p \rightarrow 0$ . This means all the energy is kinetic

$E_T = E_k = \frac{1}{2}mv_f^2$ , so

$v_f = \sqrt{2E_T/m} =\sqrt{2 *6.698 \times 10^{-2} /0.00263} = 7.137 m/s$ (mass in Kg).

That would be the velocity when the second charge moves infinitely far from the origin.

For the second part we have that $v = v_f/2$ , so kinetic energy is

$E_k = \frac{1}{2}mv^2 = 1.674 \times 10^{-2} Nm$

and potential energy is

$E_p = E_T - E_k = 6.698 \times 10^{-2} - 1.674 \times 10^{-2} = 5.024 \times 10^{-2} Nm$

so the distance is

$r = \frac{k q_1 q_2}{E_p} = 1.832\ m$

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