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Amanda [17]
3 years ago
5

A rod 14.0 cm long is uniformly charged and has a total charge of 222.0 mc. determine (a) the magnitude and (b) the direction of

the electric field along the axis of the rod at a point 36.0 cm from its center
Physics
1 answer:
Yuri [45]3 years ago
6 0

Electric field due to a charged rod along its axis is given by

E = \frac{kQ}{(L+r)(r)}

here we know that

L = 14 cm

r = distance from end of rod

r = 36 - 7 = 29 cm

Q = 222 mC

now we will have

E = \frac{(9 \times 10^9)(222 \times 10^{-3})}{(0.29)(0.43)}

E = 1.6 \times 10^{10} N/C

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NemiM [27]

Answer:

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Explanation:

(a)

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V_1+3V_2=4V_f and making V_f the subject then

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V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

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6 0
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G is the universal gravitation constant

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1 and 2 represents moon and earth

r is  the distance

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8 0
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Archy [21]

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