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2 years ago

A man walked 50 m to the south, turned, walked 40 m to the west.

Physics

1 answer:

MaRussiya [10]2 years ago

3 0

Answer:

Direction is N 51.3° E

Explanation:

» From trigonometric ratios:

${ \tt{ \red{ \tan( \theta) = \frac{opposite \: { \blue{(south)}}}{adjacent \: { \blue{(west)}}} }}} \\ \\ { \tt{ \tan( \theta) = \frac{50}{40} = \frac{5}{4} }} \\ \\ { \tt{ \theta = { \tan }^{ - 1} ( \frac{5}{4}) }} \\ \\ { \tt{ \theta = 51.3 \degree}} \\ \\ { \boxed{ \tt{ \: direction : \:{ \green{N }}\: 51.3 \degree \: { \green{ E }}}}}$

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3 years ago

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Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

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$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

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3 years ago

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Answer:

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