Answer:
13.4 x 10 raise to power -19 C
Explanation:
. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m
. The uniform electric field is E = 214 N/M
, The decrease in electrical potential energy is
d(P.E) = 51.63 x 10 raise to power -19 J
Let the magnitude of the charge of the moving particle be q
which is given by the equation
d(P.E) =qEd
51.63 x 10 power -19 = q(214)(0.018)
51.63 x 10 power -19 =3.852q
by making q the formular,
q = 13.4 x 10 power -19 C
Explanation:
We have,
Mass of an object is 0.5 kg
Force constant of the spring is 157 N/m
The object is released from rest when the spring is compressed 0.19 m.
(A) The force acting on the object is given by :
F = kx

(B) The force is simply given by :
F = ma
a is acceleration at that instant
