A 85.0 kg person holds out his arms so that his hands are 1.80m apart. Typically
ID: 1999537 • Letter: A
Question
A 85.0 kg person holds out his arms so that his hands are 1.80m apart. Typically, a person's hand makes up about 1.00 % of his or her body weight. For round numbers, we shall assume that all the weight of each hand is due to the calcium in the bones, and we shall treat the hands as point charges. One mole of Ca contains 40.18 g, and each atom has 20 protons and 20 electrons. Suppose that only 2.00 % of the positive charges in each hand were unbalanced by negative charge.
a) How many Ca atoms does each hand contain?
N =
b) How many coulombs of unbalanced charge does each hand contain?
Q =
c) What force would the person's arms have to exert on his hands to prevent them from flying off?
F =
d) Does it seem likely that his arms are capable of exerting such a force?
Explanation / Answer
weight of person = 85 Kg
weight of hands = 2* 1% of 85 = 1.7 Kg
a.) total no of Ca atom in each hnad = no of moles of Ca * 6.0233*10^23 atom
= 0.85/0.04018 * 6.0233*10^23 atom
=1.31*10^25 atoms
b.)No of unbalanced charge each hand contain = 2% 0f 1.31*10^25 atoms
= 2.62 *10^23 atoms
c.)force would the person's arms have to exert on his hands to prevent them from flying off is mg
= 2*0.85*9.81 = 16.677 N
d.)yes, it seem likely that his arms are capable of exerting such a force
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