By Newton's second law, the net force on the object acting parallel to the surface is
∑ F = F[applied] - F[friction] = (8k g) (3 m/s²)
If F[applied] = 30 N, then
30 N - F[friction] = 24 N ⇒ F[friction] = 6 N
so the answer is B.
Answer:
3984875 N/C
Explanation:
Applying,
E = kq/r².................. Equation 1
Where E = Electric field, q = charge, r = distance, k = coulomb's constant.
From the question,
Given: q = 7.10 mC = 0.0071 C, r = 4.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 1
E = 0.0071(8.98×10⁹)/4²
E = 3984875 N/C
Hence the electric field is 3984875 N/C
D
Its D i think because it doesnt refer to any other things
