The magnitude of the electric force on the charge is 5 N.
<h3>Magnitude of force on the charge</h3>
The magnitude of force on the charge is calculated as follows;
F = Eq
where;
- E is electric field
- q is magnitude of the charge
F = 100 N/C x 0.05 C
F = 5 N
Thus, the magnitude of the electric force on the charge is 5 N.
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Answer:
s = it+1/2 at²
s= 8×3+1/2 (10)(3)²
s = 24+45
s= 69
the object was thrown from a height of 69 meters
The statements of both students are incorrect.
-- Electrical power, just like mechanical power, is expressed in units of watts.
-- 'Coulomb' is the unit of electrical charge.
-- '400 k ohms' means 400,000 ohms of resistance.
-- 'Volt' is the unit of electromotive force (or potential difference).
There are no 'following statements'.
All in all, a very disappointing question.
