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prisoha [69]
3 years ago
15

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

to measure the charge on small droplets of oil by suspending them between a pair of electrically charged plates. Which of the values are possible charges of those oil droplets? − 8.0 × 10 − 19 C − 3.2 × 10 − 19 C − 1.2 × 10 − 19 C − 5.6 × 10 − 19 C − 4.8 × 10 − 19 C − 9.4 × 10 − 19 C
Physics
1 answer:
vovangra [49]3 years ago
5 0

Answer:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

q_e=-1.6\cdot 10^{-19}\ C

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 5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C

Let's test the possible charges listed in the question:

-8.0 \times 10 ^{-19 }. We have just found it's a possible charge of a particle

-3.2 \times 10 ^{-19 }. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

-1.2 \times 10 ^{-19 } 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

-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 } cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge -4.8 \times 10 ^{-19 }\ C 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:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

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Oksanka [162]

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The length of the object would shrink to zero which is not possible.

Explanation:

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The Lorentz factor is given as:

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4 0
3 years ago
A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi
bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

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rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

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4 0
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Answer:

1) Addition of a catalyst

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