Bell, Evan – Homework 1 – Due: Jan 19 2005, 11:00 pm – Inst: Furlan
1
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printout
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have
34
questions.
Multiplechoice questions may continue on
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
Assume the Sun is mostly composed of hy
drogen atoms.
Assume the mass of the Sun
is 1
.
54
×
10
30
kg, and the mass of a hydrogen
atom is 1
.
672
×
10

27
kg.
How many atoms are there in the Sun?
Correct answer: 9
.
21053
×
10
56
atoms.
Explanation:
Basic Concepts:
Elementary Arithmetic
N
atoms
=
M
sun
M
atom
=
1
.
54
×
10
30
kg
1
.
672
×
10

27
kg
= 9
.
21053
×
10
56
atoms
.
002
(part 1 of 1) 10 points
The base of a pyramid covers an area of
19
.
7 acres (1 acre = 43
,
560 ft
2
) and has a
height of 69 ft.
If the volume of a pyramid is given by the
expression
V
=
1
3
b h
, where
b
is the area of
the base and
h
is the height, find the volume
of this pyramid in cubic meters.
Correct answer: 558891 m
3
.
Explanation:
Given :
b
= 19
.
7 acres
and
h
= 69 ft
.
V
=
1
3
b h
=
1
3
(19
.
7 acres) (69 ft)
·
43560
ft
2
acre
¶
‡
0
.
3048
m
ft
·
3
=
558891 m
3
.
003
(part 1 of 3) 10 points
Consider the triangle shown in the figure.
7
.
3 m
13 m
φ
θ
What is the length of the unknown side?
Correct answer: 10
.
7569 m.
Explanation:
Given :
c
= 13 m
and
a
= 7
.
3 m
.
Applying the Pythagorean Theorem,
b
=
p
c
2

a
2
=
q
(13 m)
2

(7
.
3 m)
2
=
10
.
7569 m
004
(part 2 of 3) 10 points
Find the tangent of
θ
.
Correct answer: 0
.
678637 .
Explanation:
tan
θ
=
opposite
adjacent
=
7
.
3 m
10
.
7569 m
=
0
.
678637
.
005
(part 3 of 3) 10 points
Find the sin of
φ
.
Correct answer: 0
.
827451 .
Explanation:
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Bell, Evan – Homework 1 – Due: Jan 19 2005, 11:00 pm – Inst: Furlan
2
sin
φ
=
opposite
hypotenuse
=
10
.
7569 m
13 m
=
0
.
827451
.
006
(part 1 of 1) 10 points
A car travels along a straight stretch of road.
It proceeds for 12
.
4 mi at 52 mi
/
h,
then
24
.
2 mi at 43 mi
/
h, and finally 41 mi at
35
.
7 mi
/
h.
What is the car’s average velocity during
the entire trip?
Correct answer: 39
.
8008 mi
/
h.
Explanation:
The total distance the car traveled is
Δ
d
=
d
A
+
d
B
+
d
C
= 77
.
6 mi
The total time the car spent on the road is
Δ
t
=
d
A
v
A
+
d
B
v
B
+
d
C
v
C
=
12
.
4 mi
52 mi
/
h
+
24
.
2 mi
43 mi
/
h
+
41 mi
35
.
7 mi
/
h
= 1
.
94971 h
.
Hence the average velocity is
v
=
Δ
d
Δ
t
=
77
.
6 mi
1
.
94971 h
= 39
.
8008 mi
/
h
.
007
(part 1 of 1) 10 points
Assume:
1
.
28 km
/
lap.
A race car is one lap behind the lead race
car when the lead car has 52 laps to go in a
race.
If the speed of the lead car is 55
.
9 m
/
s,
what must be the average speed of the second
car to catch the lead car just before the end
of the race (
i. e.,
right at the finish line)?
Correct answer: 56
.
975 m
/
s.
Explanation:
Basic Concepts
Constant Velocity
Solution
The time needed for the lead car to end the
race is
t
=
n
v
=
(52 laps)
(55
.
9 m
/
s)
=
(52 laps) (1
.
28 km
/
lap)
(55
.
9 m
/
s)
1000 m
1 km
¶
= 1190
.
7 s
.
In the same period of time, the trailing car
must go 1 lap more than the lead car to catch
up; thus its average speed must be
v
avg
=
m
t
=
(53 laps)
(1190
.
7 s)
=
(53 laps) (1
.
28 km
/
lap)
(1190
.
7 s)
1000 m
1 km
¶
= 56
.
975 m
/
s
.
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 Spring '07
 GALLIS,MICHAELRO
 Acceleration, Work, Velocity, Correct Answer, m/s

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