Chp. 12 Homework pg. 292-292. Total 26 points.
Vocab. #1-15,17-18
Concepts #1-5,7,10,12
Problems #2-3,5-8,10,12.
Vocab.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
17.
18.
physical properties
chemical properties
amorphous
brittleness
crystalline
ductility
elasticity
hardness
malleability
strength
tensile strength
thermal expansion
viscosity
fluid
pressure
buoyancy
Archimede’s Principle
Concepts
1.
A material is most dense as a solid and least dense as a gas. The exception is water
Water’s solid state is less dense than it’s liquid state.
2.
Physical properties can be observed directly, without changing the substance into another
substance. Example: melting and boiling point. Chemical properties can only be observed when
a substance is changed into another substance. Example: iron reacting with oxygen to form iron
oxide (rust).
3.
Density = mass / volume. Density depends on the mass of each atom that make up the
material and how tightly packed those atoms or molecules are.
4.
Amorphous – atoms arranged randomly. Crystalline solid – arranged in orderly pattern.
5.
a.
b.
c.
d.
e.
f.
P
C
P
P
P
C
7.
a.
b.
c.
2. Ductility
3. Elasticity
1. Brittleness
10.
Air flow over the top of the curved wing moves faster than the air below the wing,
causing lower pressure on the top than the bottom. This pushes the wing up into the air.
12.
Pascal’s principle states that the pressure applied to an incompressible fluid in a closed
container is transmitted equally to all parts of the fluid. A force applied to a small area (small
hydraulic cylinder) will produce a larger force in a cylinder with a larger area.
13.
The marble is more dense than water.
16.
A steel ship has a lower average density than water.
17.
The buoyant force is the same because they both have the same volume (both displace the
same volume of water).
Problems
2.
The platinum atoms are packed more tightly together than the copper atoms.
3.
P = F/A = 810 N / (9 m2) = 90 N/m2 = 90 Pa
5.
P = F/A  A = F / P = 4000/35 = 114.3 in2 total, so 28.6 in2 per tire
6.
Pressure stays the same in the smaller cylinder as the larger cylinder. So, first, calculate
the pressure in the input cylinder: P = F/A = 15 / 0.1 = 150 Pa
Use that pressure to find the output force: F = P x A = (150 Pa) x (5 m2) = 750 N
7.
You need to find the density of air (pg. 36) to determine the mass of 6000 mL.
6000 mL x (0.001 g / mL) = 6 g
Buoyant force is equal to the weight of a fluid displaced, and the weight of matter on Earth is
due to gravity, and gravity “pulls” on matter with a force of 9.8 N/kg. So convert 6 g to kg.
6 g = 0.006 kg
So the balloon displaced 0.006 kg of air, which has a weight of
W = (0.006 kg)(9.8 N/kg) = 0.059 N
This is the buoyant force “pushing” upward on the balloon, 0.059 N.
8.
45 N – 22 N = 23 N
10.
6.5 N – 5 N = 1.5 N
12.
less so the weight pushing down of the object would not “overcome” the buoyancy force
pushing up on the object.