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faust18 [17]
3 years ago
12

7. The uranium nucleus contains a charge 92 times that of the proton. If a proton is shot at the nucleus, how large a repulsive

force does the proton experience due to the nucleus when it is 1×10-11m from the nucleus? a) 1.1 ×10-4N b) 2.1 ×10-4N c) 1.1 ×10-2N d) 2.1 ×10-2N
Physics
1 answer:
umka21 [38]3 years ago
6 0

Answer:

2.12 x 10^-4 N

Explanation:

charge on uranium nucleus, Q = 92 e

charge on proton, q = e

distance, d = 1 x 10^-11 m

The force between the two charged particles is given by

F = \frac{KQq}{d^{2}}

F = \frac{K\times 92 e \times e}{10^{-22}}

e = 1.6 x 10^-19 C

F = \frac{9\times 10^{9}\times 92\times 1.6\times 10^{-19} \times 1.6\times 10^{-19}}{10^{-22}}

F = 2.12 x 10^-4 N

Thus, the force between the proton and the nucleus of Uranium is 2.12 x 10^-4 N.

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Someone please help me with finding the resistance of these circuits! I've been asking for an hour now. I will give brainliest i
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\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} ...\\\frac{1}{R_T} = \frac{1}{6.0} +\frac{1}{12} +\frac{1}{36}+\frac{1}{18} \\\frac{1}{R_T} = \frac{1}{3} \\R_T=3

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\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} ...\\\frac{1}{R_T} = \frac{1}{10} +\frac{1}{2} +\frac{1}{1} ...\\\frac{1}{R_T} =1.6\\R_T=\frac{1}{1.6}

Therefore, the total resistance in the third circuit is \frac{1}{1.6} kΩ, or 0.625 kΩ.

I hope this helps!

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