a 75-kg person holds out his arms so that his hands are 1.7m apart. Typically a
ID: 2166803 • Letter: A
Question
a 75-kg person holds out his arms so that his hands are 1.7m apart. Typically a person's hand makes up about 1.0% for his body weight. assume all the weight of each hand is due to the calcium in the bones. one mole of Ca contains 40.18g and each atom has 20 protons and 20 electrons. suppose that only 1.0% of the positive charges in each hand were unbalanced by negative charge. a) how many Ca atoms does each hand contain? b) how many coulombs of unbalanced charge does each hand contain? c) what force would the person's arm have to exert on his hands to prevent them from flying off? d) does it seem likely that his arms are capable of exerting such a force?Explanation / Answer
Total weight of the person: 75 Kg.......
1% of total weight = 0.75 Kg= 750 grams,
......... moles of Ca present = 750 / 40.18 = 18.666 moles..
....... let A = avogadro's number = 6.022 * 10^23....
..... Part A:..........
Number of Ca atoms = 18.66 A..
..... =18.666* 6.022 * 10^23 atoms.......
.. Part b:..........
. Total number of protons=20 in each atom * number of atoms
. =20 * 18.666* 6.022 * 10^23 =2248.13* 10^23
Number of unbalanced atoms= 1%
=0.01*2248.13* 10^23 =2248.13* 10^21
Charge due to this 2248.13* 10^21 / 6.2 x 10^18 =362580 coulombs
Repulsion force due to unbalanced charge=
F = (1 / (4 * pi * Eo)) * (q*q) / (r^2)
substituting,
we get: =4*10^20 Newtons.
It is NOT possible to exert this much force.
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