Which is the best explanation of how the Moon formed?

A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.

Contact [7]

Answer:

A) $V_A = 11.93~V$

B) The vector definition of E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

$V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}$

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

$r = \sqrt{x^2+4^2}$

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

$dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}$

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

$V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx$

By using an integral table, this can be calculated:

$V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V$

B) The electric field can be found by a similar approach, but a different formula:

$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r$

Let's apply this formula to the infinitesimal part we have chosen.

$dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)$

By the geometry sine and cosine terms can be found:

$\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}$

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

$E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx = 2.41(\^y)$

So, the final E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

The magnitude of the E-field is

E = 2.66 N/m

6 0

3 years ago

Using only one management style with all people is the most effective leadership technique for any organization.

Softa [21]

the answer for this is false

4 0

3 years ago

The very strong source of radio waves at the center of our galaxy is called.

Alika [10]

The answer is Sagittarius A, a very large black hole.

4 0

3 years ago

A student is bicycling north along Main Street to school. Another student is timing the bicycling student in order to determine

MatroZZZ [7]

The average velocity is -4.17 m/s

Explanation:

The average velocity of a body is given by:

$v=\frac{d}{t}$

where

d is the displacement of the body

t is the time elapsed

For the student in this problem, we have:

Initial position: $x_i = 450 m$

Final position: $x_f = 200 m$

So the displacement is

$d=x_f -x_i = 200 - 450 = -250 m$

The time elapsed is

t = 60 s

Therefore, the average velocity is

$v=\frac{-250}{60}=-4.17 m/s$

Where the negative sign means the student is moving towards the origin.

Learn more about average speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

8 0

4 years ago

Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in se

just olya [345]

Answer:

1.61ohms and 4.39ohms

Explanation:

According to ohm's law which States that the current (I) passing through a metallic conductor at constant temperature is directly proportional to the potential difference (V) across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the current

Rt is the effective resistance

Let the resistances be R and r

When the resistors are connected in series to a 12.0-V battery and the current from the battery is 2.00 A, the equation becomes;

12 = 2(R+r)

Rt = R+r (connection in series)

6 = R+r ...(1)

If the resistors are connected in parallel to the battery and the total current from the battery is 10.2 A, the equation will become;

12 = 10.2(1/R+1/r)

Since 1/Rt = 1/R+1/r (parallel connection)

Rt = R×r/R+r

12 = 10.2(Rr/R+r)

12(R+r) = 10.2Rr ... (2)

Solving equation 1 and 2 simultaneously to get the resistances. From (1), R = 6-r...(3)

Substituting equation 3 into 2 we have;

12{(6-r)+r} = 10.2(6-r)r

12(6-r+r) = 10.2(6r-r²)

72 = 10.2(6r-r²)

36 = 5.1(6r-r²)

36 = 30.6r-5.1r²

5.1r²-30.6r +36 =

r = 30.6±√30.6²-4(5.1)(36)/2(5.1)

r = 30.6±√936.36-734.4/10.2

r = 30.6±√201.96/10.2

r = 30.6±14.2/10.2

r = 44.8/10.2 and r = 16.4/10.2

r = 4.39 and 1.61ohms

Since R+r = 6

R+1.61 = 6

R = 6-1.61

R = 4.39ohms

Therefore the resistances are 1.61ohms and 4.39ohms

5 0

3 years ago