A proton moves through a region containing a uniform electric field given by E = 32.0 ĵ V/m and a uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T. Determine the acceleration of the proton when it has a velocity v = 230 î m/s.

The acceleration of the proton is (-5.748 x 10⁹ j + 6.61 x 10⁹ k) m/s²

Explanation:

Lorentz force acting on the proton is given as;

F = ma = qE + qvB

ma = q(E + v x B)

where;

a is the acceleration of the proton

q is the charge of the proton

m is the mass of the proton

E is the electric field strength

v is the velocity of the proton

B is magnetic field strength

First, find the cross product of V  and B

Then, add E

E + V X B = (32.0 ĵ ) + (-92j + 69 k)

                = -60 j + 69 k

Finally, substitute in this value back to the main equation and calculate 'a'

Thus, the acceleration of the proton is  (-5.748 x 10⁹ j + 6.61 x 10⁹ k) m/s²

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000