Would the mass density of an object be the same if the object were on the moon rather
than on the earth? Would the weight be the same?
The mass would be the same, since mass is measure of the amount of material present
in the object and is independent of the gravitational acceleration exerted on the object.
The weight would be different. On the Moon, the object would weight approximately
1/6th of its weight on Earth. This is because the acceleration due to gravity on the
Moon is much smaller than that on Earth.
A steel column in a building has a cross-sectional area of 3500 cm^2 and supports a
weight of 2.5 x 10^5 N. Find the stress of the column.
s=
P 2.5 ´10 5 N
=
= 714,286 Pa
A
0.35 m 2
Each vertical steel column of an office building supports a weight of 1.30 x 10^5 N and is
compressed to 5.90 x 10^(-3) cm.
Find the compression in each column if a weight of 5.50 x 10^5 N is supported.
If the compression of each steel column is 0.0710 cm, what weight is supported by each
column?
The shortening of the column is given by:
d=
PL
EA
which can be rearranged as:
d
P
=
L
EA
L
is a constant (assuming that all of the columns have the same
EA
length, cross-sectional area, and are made from the same material), we can write:
Since the quantity
d1
P1
=
d2
P2
When the supported weight is 5.5 x 105 N, the compression is:
5.9 ´ 10 -3 cm
d2
=
5
1.2 ´ 10 N
5.5 ´ 10 5 N
æ 5.9 ´ 10 -3 cm ö
è 1.2 ´ 10 5 N ÷ø
d 2 = ( 5.5 ´ 10 5 N ) ç
d 2 = 2.70 ´ 10 -2 cm
When the compression is 0.0710 cm (7.1 x 10-2 cm), the supported force is:
5.9 ´ 10 -3 cm 7.1 ´ 10 -2 cm
=
1.2 ´ 10 5 N
P2
æ 1.2 ´ 10 5 N ö
P2 = ( 7.1 ´ 10 -2 cm ) ç
è 5.9 ´ 10 -3 cm ÷ø
P2 = 1, 444,068 N
The specific gravity of an unknown substance is 0.80. Will it float on or sink in gasoline?
The specific gravity of gasoline is 0.739, which indicates that the unknown substance
is more dense than the gasoline. As a result, the unknown substance will sink in
gasoline.
In a deep dive, a whale is appreciably compressed by the pressure of the surrounding
water. What happens to the whale’s density?
Density is equal to mass divided by volume. The compression will reduce the whale’s
volume, but will not change its mass. As a result, the density of the whale will
increase as a result of the compression.
Would it be slightly more difficult to draw soda through a straw at sea level than on top
of a very high mountain? Explain.
At the top of a very high mountain, the acceleration due to gravity will be slightly
less, due to the increased distance from the center of the Earth, compared to being at
sea level. As a result, the weight of the liquid being drawn through the straw will be
slightly decreased, and less work will need to be done to lift the mass of liquid
through the straw. Therefore, it would be slightly more difficult to draw the soda
through the straw at sea level than at the top of a very high mountain.
When a steadily flowing gas flows from a larger diameter pipe to a smaller diameter pipe,
what happens to the following?
Its speed.
Its pressure
The spacing between its streamlines
The volumetric flow rate of the gas (Q = VA) must remain the same. Since the crosssectional area (A) decreases, there must be a corresponding increase in V in order for
Q to remain unchanged. The speed will therefore increase.
Because the gas is flowing faster, less pressure is exerted by it on the walls of the
pipe.
Because the gas is flowing faster in the smaller pipe, the spacing between the
streamlines will be reduced (streamlines will be closer together).
The large piston in a hydraulic lift has a radius of 250 cm^2. What force must be applied
to the small piston with a radius of 25 cm^2 in order to raise a car of mass 1500 kg?
The pressure exerted on both pistons must be equal, which gives:
Psmall Plarge
=
Asmall Alarge
Substituting the given values then gives:
(1500 kg ) ( 9.81m / s
Psmall
=
2
25 cm
250 cm 2
)
( 25 cm ) (1500 kg ) ( 9.81m / s )
=
2
Psmall
2
Psmall = 1471.5 N
2
250 cm 2
Water flows through a 13.0 cm diameter fire hose at a rate of 4.53 m/s.
What is the rate of flow through the hose in L/min?
How many liters pass through the hose in 25.0 min?
The rate of flow through the hose is:
2 æ 60 s ö æ 1000 L ö
æpö
Q = VA = ( 4.53m / s ) ç ÷ ( 0.13m ) ç
= 3,607.66 L / min
è 4ø
è 1min ÷ø çè 1m 3 ÷ø
In 25 minutes, the volume of water passed through the hose will be:
( 3,607.66 L / min)( 25min) = 90,191.50 L
Suppose you cut a small gap in a metal ring. If you heat the ring, will the gap become
wider or narrower?
The gap will become wider because the entire ring will become larger as the metal
expands. (The metal will not expand just to close the gap. While the heating causes
the cut ends to expand toward each other and close the gap, the metal on the opposite
side of the ring expands to move the two cut ends farther apart.)
The temperature of a 1 m long aluminum rod is 20° C. If the temperature is increased to
70° C what is the length of the rod if the coefficient of linear expansion for aluminum is
2.3x10^(-5)/° C.
The new length will be:
∆ L = a L1 ∆ T
∆ L = L2 - L1
L2 - L1 = a L1 ∆ T
L2 = L1 + a L1 ∆ T
L2 = 1m + ( 2.3 ´ 10 -5 /º C ) (1m ) ( 70º C - 20º C )
L2 = 1.00115 m
The specific heat capacity of copper is 0.092 C/gm/° C. Show that the amount of heat
needed to raise the temperature of a 10-gram piece of copper from 0° C to 100° C is 92
calories. How does this compare with the heat needed to raise the temperature of the
same mass of water through the same difference?
The heat required to raise the temperature of the copper is:
∆ H = c p * m *∆ T
æ
cal ö
∆ H = ç 0.092
(10 g ) (100º C - 0º C )
g º C ÷ø
è
∆ H = 92 cal
The heat required to raise the temperature of 10g of water from 0ºC to 100ºC is:
∆ H = c p * m *∆ T
æ 1cal ö
∆H =ç
(10 g ) (100º C - 0º C )
è gº C ÷ø
∆ H = 1000 cal
It takes significantly more heat (almost 11 times as much) to heat the water.
A liter of water is used to cool electronics. If 1,000 Joules of heat is given off by the
electronics, by what temperature does the water increase? One liter of water has a mass of
1 kg and a specific heat of 4.184 Joules/(g C°)
∆ H = c p * m *∆ T
∆T =
∆H
cp * m
∆T =
1000 J
( 4.184 J / gº C ) (1000g )
∆ T = 0.239 º C
The water temperature will increase by 0.239 ºC.