Answer: 168.75 N
Explanation:
first, let's convert microcoulombs to coulombs
q1 = 1e-4 C
q2 = 3e-5 C
r = 0.4 m
then use the equation Fe =
plug in values --> F = (9e9*1e-4*3e-5)/(0.4)^2
F = 168.75 N
At resonance, a driven rlcrlc circuit has vcvc = 5. 0 vv , vrvr = 8. 0 vv , and vlvl = 5. 0 vv having a peak voltage across the entire circuit
For any voltage waveform, the peak voltage is the highest point or voltage value. When Pulse Width Modulation (PWM) devices, like variable frequency drives, are added to a power system with a peak voltage that is equal to the square root of two times the RMS voltage, a power quality problem results.
Find the peak voltage, for instance, if the RMS voltage is 85 V. The average voltage and maximum voltage of AC power coming from the wall are both about 110 V. Therefore, the real peak voltage is 120/0.707 = 170 V. The sinusoid's amplitude is divided in half by this. Peak-to-peak voltage (Vp-p), also known as the total amplitude, is 340 V, or twice the peak voltage.
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F = ma
4 = 40a
a = 0.1 m/s
v = u + at
-10 = 65 - 0.1t
t = 750s = 12.5 min
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