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miv72 [106K]
3 years ago
7

Diego was asked to plot these points: (-50, 0), (150, 100), (200, -100), (350, 50), (-250, 0). What interval could he use for ea

ch axis? Explain your reasoning.
Mathematics
1 answer:
ivolga24 [154]3 years ago
5 0

Answer:

Interval of 50 on both axis

Step-by-step explanation:

Given

(x_1,y_1) = (-50,0)

(x_2,y_2) = (150,100)

(x_3,y_3) = (200,-100)

(x_4,y_4) = (350,50)

(x_5,y_5) = (-250,0)

There are several ways to do this, but I will use the observation method, since the dataset is small.

Considering the x-coordinates

x = \{-50,150,200,350,-250}

Each element of the data set is a multiple of 50.

Hence, an interval of 50 can be used on the x-axis

Considering the y-coordinates

y = \{0,100,-100,50,0\}

Each element of the data set is a multiple of 50.

Hence, an interval of 50 can be used on the y-axis

<em>So, an interval of 50 can be used on both axes</em>

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

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

#SPJ4

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Step-by-step explanation:

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