Answer:
The displacement from point B to point E is 25.0 m left
F = 750 N (Force)
d = 10 m (displacement
)
t = 25 s (time)
L = ? (Mechanical work
) = (Energy)
P = ? (Power)
Solve:
L = F × d = 750 × 10 = 7500 Joules
P = L / t = 7500 / 25 = 300 Watts
The strength of the electric field is 5 N/C
Explanation:
The magnitude of the electric field produced by a single-point charge is given by:
where
is the Coulomb's constant
Q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the charge producing the field
r = 100 m is the distance from the charge at which we want to calculate the field
Substituting into the equation, we find the s trength of the electric field:
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