A 67-kg base runner begins his slide into second base when he is moving at a speed of 3.6 m/s. The coefficient of friction betwe
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Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula
E = 1461.95 N/C
c) The electric field E is calculated as:
E = 239.76 N/C
Answer:
F = 8.23 × 10⁻⁸N
Explanation:
F= kq²/r²
k= 9×10⁹Nm²/C²
q= 1.6×10⁻¹⁹C
r= 5.29×10⁻¹¹m
F= 9×10⁹× (1.6×10⁻¹⁹)²/ (5.29×10⁻¹¹)²
F = 8.23 × 10⁻⁸N
