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Thepotemich [5.8K]
2 years ago
14

Help, please

Physics
1 answer:
kvv77 [185]2 years ago
5 0

Answer:

<em>I </em><em>am</em><em> </em><em>going</em><em> to</em><em> </em><em>answer</em><em> </em><em>the</em><em> </em><em>questions</em><em>,</em><em> according</em><em> to</em><em> the</em><em> </em><em>num</em><em>bers</em><em> </em><em>of</em><em> </em><em>empty</em><em> </em><em>spaces </em><em>to</em><em> </em><em>fill</em><em>,</em><em>1</em><em>.</em><em>e</em><em>l</em><em>e</em><em>c</em><em>t</em><em>r</em><em>i</em><em>c</em><em> </em><em>charge</em><em>,</em><em>2</em><em>.</em><em>f</em><em>o</em><em>r</em><em>c</em><em>e</em><em>,</em><em>3</em><em>.</em><em>f</em><em>i</em><em>e</em><em>l</em><em>d</em><em> </em><em>lines</em><em>,</em><em>4</em><em>.</em><em>n</em><em>e</em><em>g</em><em>a</em><em>t</em><em>i</em><em>v</em><em>e</em><em>s</em><em>,</em><em>5</em><em>.</em><em>p</em><em>o</em><em>s</em><em>i</em><em>t</em><em>i</em><em>v</em><em>e</em><em>,</em><em>6</em><em>.</em><em>a</em><em>f</em><em>f</em><em>e</em><em>c</em><em>t</em><em>e</em><em>d</em><em>,</em><em>7</em><em>.</em><em>a</em><em>t</em><em>t</em><em>r</em><em>a</em><em>c</em><em>t</em><em>,</em><em>8</em><em>.</em><em>r</em><em>e</em><em>p</em><em>e</em><em>l</em><em>.</em>

Explanation:

if u read it filling in the spaces using this answers,u will understand

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Determine the magnitude and direction of the resultant velocity of 75.0 m/s. 25.0 east of north, and 100.0 m/s, 25.0 east of sou
aleksandr82 [10.1K]

Answer:

77.35 m / s

Ф = -17° from + X axis or 343° from + X axis

Explanation:

v1 = 75 m/s 25° east of north

v2 = 100 m/s  25° east of south

Write the velocities in vector form ,we get

\overrightarrow{v_{1}}=75\left ( Sin25\widehat{i} +Cos25\widehat{j}\right )=31.7\widehat{i}+67.97\widehat{j}

\overrightarrow{v_{1}}=100\left ( Sin25\widehat{i} -Cos25\widehat{j}\right )=42.26\widehat{i}-90.63\widehat{j}

Now add the velocity vectors to get the resultant of the velocities.

\overrightarrow{v}=\overrightarrow{v_{1}}+\overrightarrow{v_{2}}

\overrightarrow{v}=\left (31.7+42.26  \right )\widehat{i}+\left ( 67.97- 90.63 \right )\widehat{j}

\overrightarrow{v}=73.96\widehat{i}-22.66\widehat{j}

magnitude of resultant velocity is \sqrt{\left ( 73.96 \right )^{2}+\left ( -22.66 \right )^{2}}

  = 77.35 m / s

The direction is Ф from X axis

tan\phi =\frac{-22.66}{73.96}=-0.306

Ф = -17° from + X axis or 343° from + X axis

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