Answer:
The charge on the ball bearing 4.507 × 10^-8 C
Explanation:
From Coulomb's law
F = kq1q2/r²
make q2 the subject
q2 = Fr²/kq1
q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)
q2 = 4.507 × 10^-8 C
The answer is D. i got all the other ones wrong ._.
In scientific notation", that number would be written as
6.81 x 10⁻⁴ .
It's important because you have to know when to make it go faster or else you might not be able to go upside down or in a circle
Both planets are similar in shape and have a rocky surface. Not sure about the phases though
