Answer:
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Explanation:
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Answer:10842.33m/s
Explanation:
F=qvBsine
V=f/(qBsine)
V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)
V=10842.33m/s
Answer:

Explanation:
<u>Charge of an Electron</u>
Since Robert Millikan determined the charge of a single electron is

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 
Let's test the possible charges listed in the question:
. We have just found it's a possible charge of a particle
. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets
this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge
cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6
Finally, the charge
is four times the charge of the electron, so it is a possible value for the charge of an oil droplet
Summarizing, the following are the possible values for the charge of an oil droplet:

<span>The answer is, 7.44 kg*m/s</span>
The magnitude of electric field is produced by the electrons at a certain distance.
E = kQ/r²
where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2
Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m
Substitue the given:
E = </span>
E = 998.476 kN/C