Answer:
0.035 N
Explanation:
Parameters given:
Charge q1 = -3.31x10^(-7) C
Charge q2 = -5.7x10^(-7) C.
Distance between them, R = 22 cm = 0.22 m
Electrostatic force between to particles is given as:
F = (k* q1 * q2) / R²
F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²
F = 0.035 N
Answer:
1.2 seconds
Explanation:
distance = ((final speed + initial speed) * time)/2
Here given:
Solving steps:
3.8 = ((0 + 6.4) * time))/2
3.8 = 3.2(time)
time = 3.8/3.2
time = 1.1875 seconds ≈ 1.2 seconds
Explanation:
hope this answer was helpful
Answer:
the correct answer is C, E’= 4E
Explanation:
In this exercise you are asked to calculate the electric field at a given point
E = 
indicates that the field is E for r = 2m
E =
(1)
the field is requested for a distance r = 1 m
E ’= k \frac{q}{r'^2}
E ’= k q / 1
from equation 1
4E = k q
we substitute
E’= 4E
so the correct answer is C