Answer:
tell ke first ,what will happen in zero gravity?
Frequency = 1/time period = 1/0.05 = 20s^-1.
Start by facing East. Your first displacement is the vector
<em>d</em>₁ = (225 m) <em>i</em>
Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,
<em>d</em>₂ = (350 m) <em>j</em>
Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,
<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )
<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>
The net displacement is
<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃
<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>
and its magnitude is
|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m
If one of two interacting charges is doubled, the force between the charges will double.
Explanation:
The force between two charges is given by Coulomb's law
K=constant= 9 x 10⁹ N m²/C²
q1= charge on first particle
q2= charge on second particle
r= distance between the two charges
Now if the first charge is doubled,
we get 
F'= 2 F
Thus the force gets doubled.
