What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed
electric field of magnitude 700 N/C ?
1 answer:
Answer:
charge will be equal to
Explanation:
We have given mass of the particle m = 1.45 gram = 0.00145 kg
Acceleration due to gravity
Electric field E = 700 N/C
Electric force will be equal to , here q is charge and E is electric field
For particle to be stationary this force must be equal to force due to gravity , that is mg force
So qE = mg
So charge will be equal to
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i dont think anything will happen so maybe A: the wall is larger than you.
Explanation:
Newton said . . . F = m a
Divide each side by m . . . a = F / m
Acceleration = (force) / (mass)
Acceleration = (145.O N) / (40.0 kg)
<em>Acceleration = 3.625 m/s²</em>
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