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aliina [53]
3 years ago
9

A capacitor has plates of area 1.64 * 10 ^ - 3 * m ^ 2 . To create a capacitance of 2.38 * 10 ^ - 9 * F , how far apart should t

he plates be? [?]*10^ [?] m
Physics
1 answer:
lesya692 [45]3 years ago
6 0

Answer:

6.1 * 10^-6

Explanation:

Acellus

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A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
3 years ago
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