___________________ is the opposition to the flow of electric charge. (Resistance or Electric power)

Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19

sergejj [24]

The distance between the charges is 13.86 X 10⁴m

<u>Explanation:</u>

Given:

Force, F = 1.2N

Charge, q₁ = 1.602 X 10⁻¹⁹ C

k = 8.987 X 10⁹ Nm²/C²

Distance, d = ?

According to Coulomb's law:

$F = k\frac{q_1q_2}{r^2} \\\\$

Substituting the value in the formula we get:

$1.2 = 8.987X 10^9 X\frac{1.602 X 10^-^1^9 X 1.602 X 10^1^9}{r^2} \\\\1.2 = \frac{23.06 X 10^9}{r^2} \\\\r^2 = 19.22 X 10^9\\\\r = 13.86 X 10^4m$

Therefore, the distance between the charges is 13.86 X 10⁴m

5 0

3 years ago

The Cartesian coordinate of a point in the xy plane are (x,y)=(-3.50,-2.50)m. Find the poler coordinate of this point

Masja [62]

Answer:

The polar coordinate of $P(x,y) = (-3.50\,m,-2.50\,m)$ is $P (r,\theta) = (4.301\,m, 215.538^{\circ})$ .

Explanation:

Given a point in rectangular form, that is $P(x,y) = (x,y)$ , its polar form is defined by:

$P(x,y) = (r,\theta)$ (1)

Where:

$r$ - Norm, measured in meters.

$\theta$ - Direction, measured in sexagesimal degrees.

The norm of the point is determined by Pythagorean Theorem:

$r = \sqrt{x^{2}+y^{2}}$ (2)

And direction is calculated by following trigonometric relation:

$\theta = \tan^{-1} \frac{y}{x}$ (3)

If we know that $x = -3.50\,m$ and $y = -2.50\,m$ , then the components of coordinates in polar form is:

$r = \sqrt{(-3.50\,m)^{2}+(-2.50\,m)^{2}}$

$r \approx 4.301\,m$

Since $x < 0\,m$ and $y < 0\,m$ , direction is located at 3rd Quadrant. Given that tangent function has a period of 180º, we find direction by using this formula:

$\theta = 180^{\circ}+\tan^{-1} \left(\frac{-2.50\,m}{-3.50\,m} \right)$

$\theta \approx 215.538^{\circ}$

The polar coordinate of $P(x,y) = (-3.50\,m,-2.50\,m)$ is $P (r,\theta) = (4.301\,m, 215.538^{\circ})$ .

5 0

2 years ago

The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in

vivado [14]

The force the escaping gas exerts of the rocket is 10.42 N.

<h3>Force escaping gas exerts</h3>

The force the escaping gas exerts of the rocket is calculated as follows;

F = m(v - u)/t

where;

m is mass of the rocket
v is the final velocity of the rocket
u is the initial velocity of the rocket
t is time of motion

F = (0.25)(40 - 15)/0.6

F = 10.42 N

Thus, the force the escaping gas exerts of the rocket is 10.42 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

7 0

1 year ago

A satellite in orbit is not truly traveling through a vacuum.It is moving through very, very thin air. Does the resulting airfri

defon

PART A) Yes, the fact that there is a frictional force acting on the satellite generates a loss of energy due to friction. What causes satellite to diminish its orbit during its tour. In fact, many satellites have rectifier systems that allow them to position themselves and remain in their orbit for a long time to avoid being trapped by the Earth's gravity Force and fall into the atmosphere where they would probably be torn apart.

PART B) As a similarity, one could start by mentioning the structure of the two equations are similar and have their own constants who were responsible for supporting them. While the law of gravity speaks of the masses of the bodies the electrostatic law speaks of the charges of the bodies. For both the force is inversely proportional to the square of the distance that separates them.

However, the most notable difference between them is basically their statement. While one of the equations speaks about greavedad the other reflects the electromagnetic phenomena. It should be noted that the force of gravity is much weaker than the electromagnetic force and that the latter has the capacity of attraction and repulsion. While the gravitational force only that of attraction.

4 0

3 years ago

If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int

vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

$\Delta U = Q-W$