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Basile [38]
3 years ago
13

Please answer asap, i want to go draw with my friend!

Physics
2 answers:
Evgesh-ka [11]3 years ago
7 0

Answer:

50 to the left

Explanation:

subtract the two forces, whatever side has more force is the side that force will be on. now, go draw wit yo friend

Aliun [14]3 years ago
6 0

Answer:

A is the correct answer!

Explanation:

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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
The atmosphere of Neptune and Uranus have a blue color because of which gas?
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The gas that gives Neptune and Uranus its amazing blue color is methane 
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4 years ago
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I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove
Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
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</span> </span> </span> <span> Distance at Perihelion (</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is  1/2  of the orbital period.

</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>

1/2 (50%) of that is  43.9845  Earth days

The average of the aphelion and perihelion distances is

     1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
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This also happens to be 1/2 of the major axis of the elliptical orbit.


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4 years ago
A ferris wheel has a radius of 36 feet and completes one revolution every 5.4 minutes. what is the speed of the ferris wheel (in
KIM [24]
In order to find the speed per minute you will have to divide the two numbers together. You need to do 36/5.4 so the answer will be around 6.7 feet per minute. 
4 0
3 years ago
Why are the constellations in the summer sky different from those in the winter?
atroni [7]
This happens because of the earth rotating around the sun. So we see different constellations for different seasons.
4 0
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