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Elena-2011 [213]
3 years ago
7

1. What charge is stored in a 180.0-μFcapacitor when 120.0 V is applied to it?

Physics
1 answer:
marin [14]3 years ago
3 0

Answer:

So capacitance is charge divided by voltage and we can multiply both sides by V to solve for Q. So Q—the charge stored in the capacitor—is the capacitance multiplied by the voltage. So it's 180 times 10 to the minus 6 farads times 120 volts which is 0.0216 coulombs.

Explanation:

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F = ma
6.25 N = 0.4 kg · a
a = (6.25/0.4) m/s²      since N=kg·m/s²
a = 15.625 m/s² 

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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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