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=1.660x1027 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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 Fall '06
 Winnokur
 Physics, Work, Litre, 1.67 10 kg, 1.74 10 cm, 1.74 108 cm, 2 43560 ft

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