The electrical conductivity of a substance depends on the ability of the ____ in it to move. A) electrons B) neutrons C) protons
2 answers:
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1. The bowling ball have more potential energy as it sit on top of the building
2.The bowling ball have same potential energy and kinetic energy as it is half way through its fall
3. The bowling ball have more kinetic energy just before it hits the ground
4. The potential energy of the bowling ball as it sits on top of the building is 784J
5. The potential energy of the ball as it is half way through the fall, 20 meters high is 392J
6. The kinetic energy of the ball as it is half way through the fall is 392J
7. The kinetic energy of the ball just before it hits the ground is 784J
Explanation:
Calculating potential energy and kinetic energy for all the instances,
1. ball on top of a 40 meters tall building
Potential energy at the top of building with a height of 40m = mgh
P.E = mgh =2*9.8*40= 784J
At the top pf the building since v=0 kinetic energy is zero
2. half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second
Potential energy when it is half way through fall = mgH
where H represents new height that is equal to 20m
hence P.E=mgH=2*9.8*20= 392J
Kinetic energy of the ball is =392.04J
3. Just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second.
The potential energy of ball just before it hits the ground = mgh= 2*9.8*0=0J
kinetic energy == 784J
Answer:
atomic concentration = 2 atoms/unit cell
lattice parameter: a= 3.22 x 10⁻¹⁰ m
atomic radius: r= 1.39 x 10⁻¹⁰m
Explanation:
The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell
In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:
Vc= a³=
a³= 3.12 x 10⁻²³ m³
⇒ a = ∛(3.12 x 10⁻²³ m³) = 3.22 x 10⁻¹⁰m
Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:
a=
⇒ r= a x √3/4= (3.22 x 10⁻¹⁰ m) x √3/4 = 1.39 x 10⁻¹⁰ m