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ArbitrLikvidat [17]
3 years ago
11

object carries a charge of -8.1 µC, while another carries a charge of -2.0 µC. How many electrons must be transferred from the

first to the second object so that both objects have the same charge?
Physics
1 answer:
LenKa [72]3 years ago
8 0

Number of electrons transferred: 1.91\cdot 10^{13}

Explanation:

The charge on the first object is

Q_1 = -8.1\mu C

while the charge on the 2nd object is

Q_2=-2.0 \mu C

When they are in contact, the final charge on each object will be

Q=\frac{Q_1+Q_2}{2}=\frac{-8.1+(-2.0)}{2}=-5.05 \mu C

So, the amount of charge (electrons) transferred from the 1st object to the 2nd object is

\Delta Q = Q_1 - Q = -8.1 -(5.05)=-3.05 \mu C = -3.05\cdot 10^{-6}C

The charge of one electron is

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

Therefore, the number of electrons transferred is

N=\frac{Q}{e}=\frac{-3.05\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=1.91\cdot 10^{13}

Learn more about electrons:

brainly.com/question/2757829

#LearnwithBrainly

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Draw the vector C⃗ =A⃗ +2B⃗ .
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Answer:

A) C is 2 units towards right

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C) C is 4 units towards right and 1 unit towards down

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Drawings are given in the attachment.

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A is 4 units towards right,

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B) Since,

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B is 1 unit towards left,

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C) Since,

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A football is thrown toward a receiver with an initial speed of 16.9 m/sat an angle of 36.5◦ above the horizontal. At that insta
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Answer:

vr = 5.39 m/s

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R = vi²*Sin (2∅) / g

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⇒  R = 27.842 m

Now we can get t, using the formula:

R = vi*Cos ∅*t     ⇒   t = R / (vi*Cos ∅)

⇒   t = (27.842 m) / (16.9 m/s*Cos 36.5º) = 2.049 s

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