practice 01 – GADHIA, TEJAS – Due: Jan 22 2007, 5:00 pm
1
Question 1
part 1 of 2
10 points
The
radius
of
a
small
ball
is
around
3
.
62538 cm.
The radius of a basketball is
about 3
.
31 times larger.
What is the ratio of the surface area of the
small ball and a basketball?
Correct answer:
0
.
0912734
(tolerance
±
1
%).
Explanation:
The surface area of a sphere is 4
π R
2
, so
the area ratio is
A
small;all
A
basketball
=
4
π R
2
small ball
4
π R
2
basketball
=
R
2
small ball
R
2
basketball
=
1
3
.
31
2
=
0
.
0912734
.
Question 2
part 2 of 2
10 points
What is the ratio of their volume?
Correct answer: 0
.
027575
(tolerance
±
1 %).
Explanation:
The volume of a sphere is
4
3
π R
3
, so the
volume ratio is
V
small ball
V
basketball
=
4
3
π R
3
small ball
4
3
π R
3
basketball
=
R
3
small ball
R
3
basketball
=
1
3
.
31
3
=
0
.
027575
.
Question 3
part 1 of 1
10 points
A high fountain of water is located at the
center of a circular pool as in the figure.
A
student walks around the pool and estimates
its circumference to be 144 m.
Next, the
student stands at the edge of the pool and uses
a protractor to gauge the angle of elevation of
the top of the fountain to be 46
.
7
◦
.
How high is the fountain?
Correct answer:
24
.
3203
m (tolerance
±
1
%).
Explanation:
R
h
θ
The circumference of the pool is
C
= 2
πR
so its radius is
R
=
C
2
π
=
144 m
2
π
= 22
.
9183 m
The radius and the height of the fountain form
the adjacent and opposite sides of the angle
of elevation, respectively. Thus the tangent of
the angle can be used, and
tan
θ
=
h
R
so that
h
=
R
tan
θ
= 24
.
3203 m
Question 4
part 1 of 1
10 points
Two ants race across a table 75 cm long.
One travels at 5
.
59 cm
/
s and the other at
3
.
99999 cm
/
s.
When the first one crosses the finish line,
how far behind is the second one?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
practice 01 – GADHIA, TEJAS – Due: Jan 22 2007, 5:00 pm
2
Correct answer: 21
.
3329
cm (tolerance
±
1
%).
Explanation:
Let :
ℓ
= 75 cm
,
v
1
= 5
.
59 cm
/
s
,
and
v
2
= 3
.
99999 cm
/
s
.
You first have to compute the time it takes
the first (faster) ant to cross the finish line:
t
=
ℓ
v
1
.
Then you must compute the distance the
slower ant covers in that time
s
2
=
v
2
t
=
v
2
ℓ
v
1
.
The slower ant is
ℓ

s
2
=
ℓ

v
2
ℓ
v
1
= (75 cm)

(3
.
99999 cm
/
s) (75 cm)
(5
.
59 cm
/
s)
=
21
.
3329 cm
from the finish line when the faster one crosses
it.
Question 5
part 1 of 1
10 points
A body moving with uniform acceleration
has a velocity of 6
.
46 cm
/
s when its
x
coordi
nate is 4
.
59 cm.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Acceleration, ﬁrst cube

Click to edit the document details