Determine the magnitude and direction of the electric force on an electron in a uniform electric field of strength 2460 N/C that points due east.

30 fN ( 30* 10^-15 N) and it is pointed to to -y-axis.

Explanation: In order tosolve this question we have to usethe Coulomb force given by:

F=e*E= 1.6 *10^-19 C* 1920 N/C= 30 fN the force pints in opossite direction of the electric field due to the sign of charge of the electron.

a) electrical > light, sound, and thermal< < < <

) the mechanical energy of the firefighter moving.< < < < a) electrical > light, sound, and thermal< < < <

c) both a and b.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg