Mr. Sarver
Review Questions
6.2
The volume of a crushed loaf of bread reduced but the mass is unchanged. The density is
increased.
6.4
Pressure is force per square area.
6.6
Double the depth, double the pressure.
Saltwater is more dense than freshwater so the pressure will be equally greater.
6.8
The water flows out of the container in the direction normal to the surface of the container.
6.10
The volume of the submerged object is equal to the volume of the water displaced.
6.12
Submerged is completely immersed. And immersed is in the fluid at least a little bit. The
difference between a square and rectangle.
6.14
1L of water has a mass of 1kg and has a weight of 9.8N
6.16
Buoyant force depends on the weight of the displaced fluid and the volume of the object.
The buoyant force is the weight of the displaced fluid trying to get back into its space. The
amount of displaced fluid is determined by the volume of the object doing the displacing.
The second question is ambigous between net force and bouyant force. Question 18 suggest
net force. Net force depends on the weight of the object and its volume. I took just its
weight if given.
6.18
I believe I just said that the buoyant force is determined by the volume of the object doing
the displacing.
6.20
Wolframalpha.com is informing me that a ton is a unit of mass equal to 907.2kg. The weight
of the water displaced by that ship is 889kN. Due to the ambiguity of US Custom units like
the pound and ton, the weight of the water displaced is 100 tons. The bouyant force of water
on the ship is 889kN or 100 tons.
1
6.22
Halving the volume of any fluid doubles the density.
6.24
Boyle’s law is that pressure times volume is the equal to a constant given that the energy
(temperture) of the gas is unchanged. I can see application gas dynamics software and
weather forcasting.
6.26
F
P = FA = 1atm = 101325P a = 1cm
2
2
N
m
2
101325 m2 × 1cm × 100cm2 = 10.1325N = F = W
10.1325N
F
= m = 9.80665
m = 1.0332274527999kg
a
s2
The weight of such a column of air is 10. N and has a mass of 1.0 kg.
6.28
The weight of the mercury in a barometer is equal to the weight of the air of equal horizontal
cross section.
6.30
It would be more accurate to say that the water is pushed up the straw by the atmosphere.
Creating a vacuum in straw means that the there is nothing in the straw to push or pull.
6.32
There is 1 N of buoyant force. If the buoyant force decreases, the balloon accerates down.
If the buoyant force increases, the balloon accelerates up.
Exercises
6.3
The density of the whale increases as the whale’s volume decreases.
6.4
I’m not sure how compressible I am. Most of me is incompressible water. There is some
compressible air space in my body outside of lungs. What air there is in me can most certainly be compressed. I think there is an increase in my density, mostly in air that, but not
enough to worth noting the increase.
6.38
Changing helium to hydrogen, does not increase the bouyant force if the balloon keeps the
same volume. It rises faster because it is weighs less.
Problems
6.1
=
P = mg
A
5kg×9.8 m2
s
1cm2
N
= 49 cm
2 ×
(100cm)2
(1m)2
= 490000 mN2 = 490kP a
2
2
ρH2 O = 1 cmg 3 ×
(100cm)3
m3
3
kg
1000g
3
×
kg
= 1000 m
3
m
V = 1L × 1000L
= .001m
6kg
kg
m
ρ = V = .001m3 = 6000 m
3
The block is 6 times denser than water.
6.6
kg
kg
= .0193 cm
ρAu = 19.3 cmg 3 × 1000g
3
1kg
m
V = ρ = .0193 kg = 51.813471502591cm3
cm3
6.8
kg m
kg
ρw = 1025 m
ρv = m
= V
3
3
3
d = [m ] − displaced water
V = [m3 ] − volume vacationer
d
= .9
V
ρw d = mw = mv
ρw d
= mVv = ρV
V
kg d
kg
kg
ρw d
= 1025 m
3 V = 1025 m3 .9 = 922.5 m3
V
6.10
The weight of the displaced air is 20,000 N.
6.12
P =
F1
A1
=
F2
A2
F1 A2
= F2
A1
π(6cm)2
F1 π(2cm)2 = F1 ( 62 )2
⇒
A2
F2 = F1 A
=
1
The force is multiplied by 9.
= 9F1
4 P1 V1 = P2 V2 ⇒ PP1 V2 1 = V2
V2 = 2atm×3L
= 2L
3atm
3