The direction of an electric field is the direction of the force that the field would exert on

Aparticle with mass 1.81×10−3 kg and a charge of 1.22×10−8 c has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. what are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field b⃗ =(1.63t)i^+(0.980t)j^?

How is acceleration calculated? a. initial velocity - final velocity / timeb. initial velocity + final velocity / timec. final velocity - initial velocity / timed. final velocity / timele select the best answer from the choices provided0oood00save and exitnextmark this and return

The scalar triple product computes the magnitude m of the moment of a force vector f about a specified line. it is m = (r × f) · n, where r is the position vector from the line to the point of application of the force and n is a unit vector in the direction of the line. use matlab to compute the magnitude m for the case where f = [12, −5, 4] n, r = [−3, 5, 2] m, and n = [6, 5, −7].