Physics 1110
Problem Set 1

Note: Include
units in all problems

A. Write the
following numbers in ordinary
notation:

1)
10^{5}
cm
3) 2.56 x 10^{8 }ft^{3}

2)
10^{-5 }sec
4) 73.89 x 10^{9 }BTU’s

B. Express
55mi/H in km/H

C. Express
720 in^{2 }in ft^{2}

^{ }

D. Express
6048 in^{3 }in ft^{3}

^{ }

E. At a cost of $5.50 per square yard of
material, how
much will a person have to pay for a strip of

material 4 feet wide by 20 feet long?

F. A water
faucet has a cross sectional
area of 1.5 in^{2}. Suppose water flows from the faucet at a
speed of

6 in/sec. What volume of water
flowing from the faucet
during every second?

G. A
horizontal force of 10 lbs. is
exerted on a block in sliding it along a horizontal surface a distance
of

7.2 ft. How much work has been done by the force?

H. A vertical
force of 4.5 Newtons
is applied to a crate in order to lift it to a table 1.2 m high. How
much

work has been done by the force?

Physics 1110
__Solution to Problem Set 1__

(Note: Please do not look at this until you have
attempted
all of the previous problems in Problem Set 1.)

A. 1)
100,000cm
3) 256,000,000 ft^{3
}

2)
0.00001 sec
4) 73,890,000,000
BTU’s

## B.
1.6 km = 1mi, hence (55 mi/H)(1.6
km/1 mi) =88 km/H

## C.
144 in^{2 }= 1 ft^{2},
hence 720 in^{2}
(1 ft^{2}/ 144 in^{2}) = 5 ft^{2}

## D.
1728 in^{3 }= 1 ft^{3},
hence 6048 in^{3}
(1 ft^{3}/1728 in^{3}) = 3.5 ft^{3}

## E.
The area of the material is 4ft x 20
ft =80ft^{2}

However,
3 ft = 1yd or 9ft^{2 }= 1yd^{2}, hence 80 ft^{2}
(1yd^{2}/9ft^{2})($5.50/yd^{2})
= $48.89

##

## F.
This particular problem we have not
determined how to
do in class. However, one can deduce how to work the problem from what
is asked
for in the problem—i.e. what volume (in^{3}) is flowing per
second(s)?
Indeed the volume rate of flow can be determined by multiplying the
speed (or
velocity) times the area. Hence (6 in/sec)(1.5
in^{2})
= 9in^{3}/sec

## G.
W =Force x distance = (10 lbs)(7.2
ft) = 72 ft. lbs.

# H. W=
Force x distance = (4.5 N)(1.2m) = 5.4 joules