Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
Because the four outer planets are comprised of mostly gases give me a thanks or brainiest answer if this helps!
Answer:
Series
Explanation:
Because I listen to my science teacher
Answer:
Options A and D
Explanation:
In this question the student needs to collect these measurements in order to approximate the work done
A. The mass of the student
D. The final vertical height above the initial vertical position.
Workdone = mgh
m = mass
g = gravity = 9.8m/s²
h = vertical height between the initial and the final positions.
The vertical height has to be known as gravity only acts straight down.
