ENTA 11005 – Engineering Science
1 of 7
Assignment 1 Solutions
Question 1
(3 Marks)
(a)
We need to calculate the volume conversion from pints/pots to m
3
L
mL
cm
L
m
cm
m
m
1
1000
1000
1
1
100
1
1
3
3
3
3
×
×
×
=
mL
mL
mL
m
6
3
3
10
1
100000
100
1
1
×
=
=
×
=
∴
Knowing that 1m
3
is equivalent to 1×10
6
mL, we can determine how many pints/pots
are in 1m
3
;
For Paddy
;
No. of pints in 1m
3
4
.
1754
570
10
1
570
1
6
3
=
×
=
=
mL
mL
mL
m
For Steve
;
No. of pots in 1m
3
8
.
3508
285
10
1
285
1
6
3
=
×
=
=
mL
mL
mL
m
So Paddy will be drinking 1754 pints, and Steve will have 3509 pots throughout the
year.
(b)
To find the mass
;
V
m
=
ρ
,
or
V
m
×
=
ρ
where
ρ
is density,
m
is mass and
V
is volume.
ρ
= 1.35
kg
/
m
3
3
4
6
3
10
7
.
5
10
1
1
570
m
mL
m
mL
V
−
×
=
×
×
=
(
)
(
)
6555
.
0
10
7
.
5
10
15
.
1
4
3
=
×
×
×
=
×
=
−
V
m
ρ
The mass of one pint is 0.656
kg
, or 656
g
.
To find the weight
;
mg
W
=
, where
m
is mass and
g
is acceleration due to gravity.

ENTA 11005 – Engineering Science
2 of 7
m
=0.6555kg
g
= 9.8
ms
-2
4239
.
6
8
.
9
6555
.
0
=
×
=
W
The weight of one pint is 6.4
N
.
(c)
The mass will be exactly the same on the moon as it is on earth (or anywhere in the
universe!)
So mass is 656g.
Weight on the other hand is the force with which an object experiences due to gravity.
Because the acceleration due to gravity is much smaller on the moon than it is on
earth, the weight of the pint will be considerably less.
09385
.
1
67
.
1
6555
.
0
=
×
=
=
moon
moon
mg
W
So the weight of the pint on the moon is 1.09
N
.

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- Spring '08
- Beavan
- Mass, Orders of magnitude, Engineering Science