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3 years ago

6

Please answer! I’ll give branliest! This is needed by tomorrow! Answers would be appreciated!

1 answer:

Ilya [14]3 years ago

8 0

Answer:

Top: Ne: Day, Summer. Nw: Night, Winter.

Position 2 Top: Ne: Night, Winter. Nw: Day, Summer

Bottom P1: Se: Day, Summer. Sw:Night, Winter.

Bottom P2: Se: Night, Winter. Sw: Day, summer

Step-by-step explanation:

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Multiply the Polynomials<br> 2a2(3-5)-6a(8q2 - 2a + 4)

Amiraneli [1.4K]

Answer:

<h3>-48aq2+8a2-24a</h3>

Step-by-step explanation:

first you subtract the numbers

then multiply

then distribute

combine like terms

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3 years ago

Find the magnitude of the net electric field at the origin (created by charges q1 and q2).

Readme [11.4K]

The direction of the net magnetic field is $167.36^{\circ}$

<h3>How to find the magnitude?</h3>

$q_1=-4 \mathrm{nC} \text { at }(0.6,0.8)$

$q_2=6 \mathrm{nC}$ at (0.6,0)

$r_1=$ Distance of $q_1$ from origin $=\sqrt{0.6^2+0.8^2}$

$r_2=$ Distance of $q_2$ from origin $=0.6$

$\mathrm{k}=$ Coulomb constant $$=9$ times $10^9 \mathrm{Nm}^2 / \mathrm{C}^2$$

Electric field is given by

$&E_1=\frac{k q_1}{r_1^2} \\$

$&\Rightarrow E_1=\frac{9 \times 10^9 \times 4 \times 10^{-9}}{\sqrt{0.6^2+0.8^2}} \\$

$&\Rightarrow E_1=36 \mathrm{~N} / \mathrm{C}$

$&\theta=\tan ^{-1} \frac{0.8}{0.6} \\$

$&\Rightarrow \theta=53.13^{\circ} \\$

$&E_1=36 \cos 53.13^{\circ} \hat{i}+36 \sin 53.13^{\circ} \hat{j} \\$

$&\Rightarrow E_1=21.6 \hat{i}+28.8 \hat{j}$

$&E_2=\frac{k q_2}{r_2^2} \\$

$&\Rightarrow E_2=\frac{9 \times 10^9 \times 6 \times 10^{-9}}{0.6^2} \\$

$&\Rightarrow E_2=150 \mathrm{~N} / \mathrm{C}$

The charge $q_2$ is on the $\mathrm{x}$ axis itself and it is pointing towards the origin (left side) so the sign will be negative

$E_2=-150 \hat{i}$

Resultant electric field

$&E=E_1+E_2 \\$

$&\Rightarrow E=21.6 \hat{i}+28.8 \hat{j}+(-150 \hat{i}) \\$

$&\Rightarrow E=-128.4 \hat{i}+28.8 \hat{j}$

Magnitude of electric field is given by

$&|E|=\sqrt{(-128.4)^2+28.8^2} \\$

$&\Rightarrow|E|=121.6 \mathrm{~N} / \mathrm{C}$

Magnitude of the net electric field is $121.6 \mathrm{~N} / \mathrm{C}$

Direction is given by

$\theta=\tan ^{-1} \frac{128.4}{28.8}=77.36^{\circ}$

From the -x axis

$(90+77.36)^{\circ}=167.36^{\circ}$

The direction of the net magnetic field is $167.36^{\circ}$

To learn more about magnetic field, refer to:

brainly.com/question/26257705

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