The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2) where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.
Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span>
<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
