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yarga [219]
4 years ago
7

I need help in physics please help me

Physics
1 answer:
artcher [175]4 years ago
3 0

The first step is to represent the vectors shown in the image in Cartesian coordinates.

For the vector C we have a magnitude of 4.8 and an angle 22 ° with the axis -y (direction j)

To write this vector in Cartesian coordinates we must find its component in x (address i) and in the y axis.

x = 4.8sin(22)i\\\\y = 4.8cos(22)(-j)

So:

C = 4.8sin(22) i + 4.8cos(22)(-j)\\\\C = 1.798\ i - 4.450\ j

For Vector B we have a magnitude of 5.6 and an angle of 33 with the -x axis (-i direction)

So:

x = 5.6cos(33)(-i)\\\\y = 5.6 sin(33)(-j)

So:

B = 5.6cos(33)(-i) + 5.6sin(33)(-j)\\\\B = -4.696\ i - 3.05\ j

Finally the sum of B + C is made component by component in the following way:

F = (-4.696 +1.798)i + (-4.450 - 3.05)j\\\\F = -2.898\ i - 7.5\ j

Finally the magnitude of f is:

|F| = \sqrt{(-2,898)^2 + (-7.5)^2}

| F | = 8.04

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A 9 newtwon charge is Felt on a (1x10^-4) C charge from another charge that is 3 meters away. What is the magnitude of that othe
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The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The magnitude of that other charge is found as;

\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9  = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C

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To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

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