Pickering, Tracey – Quiz 1 – Due: Feb 15 2006, 10:00 pm – Inst: Drummond
1
This
printout
should
have
27
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
Newton’s law of universal gravitation is
F
=
G
M m
r
2
.
Here,
M
and
m
are masses and
r
is the sep
aration distance.
The dimension of force is
specified by the equation
F
=
ma
.
What are the SI units of the constant
G
?
1.
[
G
] = m
/
kg
/
s
2
2.
[
G
] = N m
3.
[
G
] = J s
/
kg
4.
[
G
] = N m
/
s
2
5.
[
G
] = m
2
/
kg
2
/
s
2
6.
[
G
] = kg
/
m
2
/
s
2
7.
[
G
] = m
3
/
kg
/
s
2
correct
8.
[
G
] = W
/
m
3
9.
[
G
] = m
2
/
kg
10.
[
G
] = m
3
/
kg
2
/
s
2
Explanation:
F
=
G
M m
r
2
so
G
=
F
r
2
M m
.
Dimensional analysis
of
G
:
kg m
s
2
¶
m
2
(kg)
2
=
m
3
(kg s
2
)
.
The common expression is [
G
] = m
3
/
kg
/
s
2
.
002
(part 1 of 1) 10 points
Given:
An acre is an area equivalent to that
of a rectangle 60
.
5 yd wide and 80 yd long.
There are 36 inches in one yard.
There are
2.54 cm in one inch.
In May 1998, forest fires in southern Mexico
and Guatemala spread smoke all the way to
Austin. Those fires consumed forest land at a
rate of 23500 acres
/
week.
On the average, how many square meters
of forest are burned down every minute?
Correct answer: 9434
.
64 m
2
/
min.
Explanation:
A conversion factor for
week
min
can be easily
derived
1 wk
7 days
·
1 day
24 hr
·
1 hr
60 min
≡
wk
min
.
A conversion factor for
m
2
acre
can also be de
rived
36 in
1 yd
·
2
.
54 cm
1 in
·
1 m
100 cm
¶
2
yd
2
acre
≡
m
2
acre
,
where
yd
2
acre
is given in the problem.
Therefore the rate
R
in square meters per
minutes is
R
=
•
m
2
/
acre
week
/
min
‚
(23500 acres
/
week)
=
•
(0
.
836127 m
2
/
yd
2
)(4840 yd
2
/
acre)
(10080 min
/
week)
‚
×
(23500 acres
/
week)
= 9434
.
64 m
2
/
min
.
003
(part 1 of 1) 10 points
The graph below shows the velocity
v
as a
function of time
t
for an object moving in a
straight line.
t
v
0
t
Q
t
R
t
S
t
P
Which of the following graphs shows the
corresponding displacement
x
as a function of
time
t
for the same time interval?
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Pickering, Tracey – Quiz 1 – Due: Feb 15 2006, 10:00 pm – Inst: Drummond
2
1.
t
x
0
t
Q
t
R
t
S
t
P
2.
t
x
0
t
Q
t
R
t
S
t
P
3.
t
x
0
t
Q
t
R
t
S
t
P
4.
t
x
0
t
Q
t
R
t
S
t
P
5.
t
x
0
t
Q
t
R
t
S
t
P
correct
6.
None of these graphs are correct.
7.
t
x
0
t
Q
t
R
t
S
t
P
8.
t
x
0
t
Q
t
R
t
S
t
P
9.
t
x
0
t
Q
t
R
t
S
t
P
Explanation:
The displacement is the integral of the ve
locity with respect to time:
~x
=
Z
~v dt .
Because the velocity increases linearly from
zero at first, then remains constant, then de
creases linearly to zero, the displacement will
increase at first proportional to time squared,
then increase linearly, and then increase pro
portional to negative time squared.
From these facts, we can obtain the correct
answer.
t
x
0
t
Q
t
R
t
S
t
P
004
(part 1 of 3) 10 points
Consider the plot below describing motion
along a straight line with an initial position of
x
0
= 10 m.
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 Fall '08
 Turner
 Force, Friction, Correct Answer, Pickering

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