Answer:
-611.32 N/C
0.43723 m
Explanation:
k = Coulomb constant =
q = Charge = -4.25 nC
r = Distance from particle = 0.25 m
Electric field is given by
The magnitude is 611.32 N/C
The electric field will point straight down as the sign is negative towards the particle.
The distance from the electric field is 1.71436 m
We have that the magnitude of the gravitational force is mathematically given as
f=6.377N
<h3>Force</h3>Question Parameters:
Earth exerts a 100 N gravitational force on a metal box.
(Mass of the earth is 6e24 kg and radius of the earth is 6.4e6m.)
Generally the equation for the Gravitational mForce is mathematically given as
f=6.377N
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Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;
where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.