Physics 113
Spring 2007
Problem Set 10 Solutions
Due to time constraints, these solutions will be brief. If you have further questions, please feel free to politely contact Your Humble Instructor at
w
Physics 113
Spring 2007
Problem Set 14 Solutions
As in previous solutions, for numerical answers the intermediate calculations
have been kept to at least nine-gure precision, with the answers rounded
Physics 113
Spring 2007
Problem Set 12 Solutions
(1a)
Exercise 8-16 Solving Equation (8.1) for the common mass m1 =
m2 = m,
m=
(1b)
F r2
=
G
(25 1066 N) (0.14 m)
= 8.6 kg.
6.67 1011 N m2 /kg2
Exercise
Physics 113
Spring 2007
Problem Set 11 Solutions
Due to time constraints, these solutions will be brief.
If you have fur-
ther questions, please feel free to politely contact Your Humble Instructor at
Physics 113
Spring 2007
Problem Set 15 Solutions
(1(a) Exercise 11-15 Take the east direction to be the -direction and the
north direction to be the -direction. The initial angular veloctiy is then in
Physics 113
Spring 2007
Problem Set 16 Solutions
(1(a) Exercise 12-16 For the situation in Figure 12.11(a), the vector sum
F1 + F2 = 0, so any force must be opposite the sum, or F3 = F F = F (+ ).
For
Physics 113
Spring 2007
Problem Set 21 Solutions
(1(a) Exercise 16-17
A classic, often seen in game shows and trivia contests. What we do is solve Equation (16.2) and the equation TF = TC simultaneous
Physics 113
Spring 2007
Problem Set 20 Solutions
(1(a) Exercise 15-18
The pressure supported by the respective columns of mercury are Hg g h.
1 mm of mercury:
13.6 103 kg/m3
9.8 m/s2 (1.00 103 m) = 13
Physics 113
Spring 2007
Problem Set 19 Solutions
(1(a)
Exercise 14-36
Take a look at the discussion at the top of Page 231, in which its explained that the
beat frequency detected is twice the frequen
Physics 113
Spring 2007
Problem Set 17 Solutions
These solutions will use, without apology, the relations
k
,
m
=
= 2f,
T=
2
1
=.
f
Intermediate numerical calcultations will not, in general, be inclu