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Mars2501 [29]
4 years ago
10

Consider the three displacement vectors

Physics
1 answer:
matrenka [14]4 years ago
4 0

Answer:

Explanation:

\overrightarrow{A} = 3\widehat{i}+3\widehat{j}

\overrightarrow{B} = \widehat{i}-4\widehat{j}

\overrightarrow{C} = -2\widehat{i}+5\widehat{j}

(a)

\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}

\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}

\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}

Magnitude of \overrightarrow{D} = \sqrt{2^{2}+4^{2}}

                                                                     = 4.47 m

Let θ be the direction of vector D

tan\theta =\frac{4}{2}

θ = 63.44°

(b)

\overrightarrow{E} = - \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}

\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}

\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}

Magnitude of \overrightarrow{E} = \sqrt{6^{2}+6^{2}}

                                                                     = 8.485 m

Let θ be the direction of vector D

tan\theta =\frac{6}{-6}

θ = 135°

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