<span>C. It is the difference in electrical potential energy between two places in an electric field.</span>
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F =
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F =
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a =
21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
-- material and thickness of the interconnecting wires;
-- number, type, and configuration of all the devices
between the in- and out-terminals of the circuit ...
the points that connect to the battery;
-- quality of the connections at the points where
circuit devices connect to wires or to each other.
Using Newton's second law of motion:
F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)
Given: Find: Formula: Solve for m:
F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2
a= 200m/s^2
Solution:
Using Eq.2
m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg