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phys207hw1.sxw

# phys207hw1.sxw - Physics 207 Homework 1 Solutions Due...

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Physics 207 Homework 1 Solutions Due 9/14/05 Assignment: Ch 1, problems 6, 10, 14, 17, 23, 30, 38, 43, 48, 50 Prob 1.6) The volume of an individual sphere is given by: V 4 3 r 3 (6.1) and the mass is given by the relation: m V (6.2) If we substitute equation 1 into 2 we find that: m 4 3 r 3 (6.3) Now, dividing equation 6.3 for the large sphere by equation 3 for the small sphere and using that the ratio of large mass to small mass is five gives the following: m large m small 4 3 r large 3 4 3 r small 3 r large 3 r small 3 5 (6.5) Multiplying by the cube of the small radius and taking the cube root gives us the answer: r large 3 r small (6.6) Prob 1.10) (a) The volume of a cube is given by the cube of the length so the mass is given by the density times the length cubed: m V L 3 7.86 g cm 3 5.00 10 6 cm 3 9.83 10 16 g 9.83 10 19 kg (10.1) (b) Note the atomic mass unit is given in the front of the textbook: u 1.660 10 27 kg (10.1) u=1.660x10-27 kg. The number of atoms is given by the ratio of masses: N m object m massunit m object 55.9 u 9.83 10 19 kg 55.9 1.660 10 27 kg 1.06 10 7 atoms (10.2) Prob 1.14) Using dimensional analysis means examining the units. First we note circumference is a measure of distance so it has dimensions of L . Then area has dimensions of L 2 and volume has dimensions of L 3 . Then we analyze the equations to see what dimensions they have. Now (i) has dimensions of

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