What must the charge (sign and magnitude) of a particle of mass 1.50 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 N/CN/C ?

q = -2.19 x 10⁻⁵ C

Explanation:

Given;

mass of the particle, m = 1.5 g = 0.0015 kg

magnitude of electric field, E = 670 N/C

Electric field is given by;

where;

q is the magnitude of the

f is the force of the charge

f = mg

Since the electric field is acting downward, the force on the charge must be acting upward. Therefore, the charge must be negative

q = -2.19 x 10⁻⁵ C

ammonium chloride & tallow

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//