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dsp73
3 years ago
14

What is the amount of charge in the electrons of 5 grams of nickel?

Physics
1 answer:
sergiy2304 [10]3 years ago
6 0
1.) Two positive charges of 4.3 mC each are separated by 0.25 m.
What is the size and type of force between the two charges? Help

2.) A positive charge of 1.32 x 10-4 C and a negative charge of 9.8 mC
are 0.015 m apart. What is the size and type of force between them? H

3.) Two electrons in an atom are separated by 1.5 x 10-10 m , the typical size
of an atom. What is the size and type of force between them? H

4.) A negative charge of 4.7 mC exerts a repulsive force of 51.0 N on a 2nd charge
0.062 m away. What is the size and polarity (pos or neg) of the 2nd charge? H

5.) A negative charge of 9.3 x 10-5 C exerts an attractive force of 37.4 N when
placed 0.080 m away from a 2nd charge. Find size and polarity of 2nd charge. H

6.) Two positive charges of 3.0 mC exert a repulsive force of 2.0 N on each other.
By what distance are they separated? H

7.) Two charges q1 and q2 are separated by a distance, d and exert a force, F1
on each other. What new force (F2) will exist if:

a) q1 is doubled. H d) d is doubled. H
b) q1 is doubled and q2 is tripled. H e) d is tripled. H
c) q1 is cut in half. H f) q2 is doubled and d is cut in half. H
8.) a) How many electrons are there in 1 C ? H
b) How many excess electrons are on a ball with a charge of -5.26 x 10-17 C ? H
c) How many excess protons are on a ball with a charge of 7.29 x 10-12 C ? H

9.) How many coulombs of negative charge does a 5 gram nickel coin have?
[takes 3 steps; do a) , b) and c) to get answer, don't round off until part c]

a) Find the number of atoms in a nickel.
Its mass is 5 g and nickel has 6.02 x 1023 atoms / 58 g. H

b) Find the number of electrons in the coin. There are 28 electrons / atom. H

c) Find how many coulombs of negative charge in a nickel. 1.6 x 10-19 C / 1 e- H

10.) A lightning bolt transfers 35 C to Earth. How many electrons are transferred? H
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m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

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