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Veseljchak [2.6K]
3 years ago
8

wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

electric field at point x = 6 m.

Physics
1 answer:
lys-0071 [83]3 years ago
5 0

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

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shusha [124]

Answer:

Acceleration, a=\dfrac{1}{8}(-i+9j)\ m/s^2

Explanation:

Initial velocity of a particle in vector form, u = (-5i - 2j) m/s

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Time taken, t = 8 seconds

We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

a=\dfrac{v-u}{t}

a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}    

a=\dfrac{(-i+9j)}{8\ s}\ m/s^2    

or

a=\dfrac{1}{8}(-i+9j)\ m/s^2

Hence, the value of acceleration vector is solved.

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Draw a simple circuit that lights up a bulb. ​
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Answer:

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2 years ago
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grandymaker [24]

Answer:

Depends.

Explanation:

Whether the object is going left or right, the speed will stay the same until friction eventually stops it. <em>However, </em>if, for example, we're talking about an object going straight before veering right, then yes, speed <em>does</em> matter. An object will normally have to speed up or slow down momentarily when changing direction to keep itself sustained on the ground.

So, honestly? It really depends on what we're talking about!

Hope this helped!

Source(s) used: None.

7 0
2 years ago
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