2 years ago

If the electric field is 100N/C at a distance of 50 cm from a point charge , what is the value of ?.

1 answer:

5 0

Answer:

$e \: = \: \frac{kq}{{r}^{2} }$

then q = e r² / k

then q = 100 * (50 ×10^-2)² / 9×10^9 = 2.777777778×10^-9 C

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first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

$d_1 = v*t_1$

$d_1 = 10*20 = 200 m$

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

$d_2 = v*t = 20*20 = 400 m$

so the net displacement is given as

$\vec d = \vec d_2 - \vec d_1$

$\vec d = 400 - 200 = 200 m$

so it will be displaced by total displacement 200 m

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3 years ago

ra1l [238]

What is that?? Please tell us

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3 years ago

vivado [14]

Strong alien you got there good luck bud you never asked a question

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3 years ago

zhannawk [14.2K]

Work done = force * distance moved (in direction of the force)

force= mass* acceleration

force=58.1N

58.1*(5.8*10^4)
=3,369,800 J

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3 years ago

olga_2 [115]

Answer:

The correct answer is Option A.

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3 years ago