Answer:
F1 = K Q1 Q2 / R1^2
F2 = K Q1 / 2 * Q2 / (2 R1)^2
F2 / F1 = 1/2 / 4 = 1/8
The new force is 5N (1/2 due to charge and 1/4 due to distance)
In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs
Answer:
d = 13 miles
Explanation:
Lets say the position of court house is origin in this case
her office is located at 4 miles west and 4 miles south of court house
so here we have coordinate of the office with respect to court house is given as
now the position of her home is located at 1 miles east and 8 miles north of the court house
so the coordinates of her home is given as
now the change in the position is given as the distance between office and home
The correct answer would be D
