Explanation:
Newton's second law simply says that the net force on an object is equal to the object's mass times its acceleration.
∑F = ma
For example, think of a game of tug-of-war, in which two teams pull on a rope in opposite directions.
If the forces are equal (balanced), then the net force is 0 N, so Newton's second law tells us that the rope's acceleration is 0 m/s².
If the forces are not equal (unbalanced), then the net force is not 0 N, and the rope will accelerate in the direction of the net force.
Well I would assume it would increase due to the increase in body movement creating more energy
Answer:
1.25 m/s
Explanation:
Given,
Mass of first ball=0.3 kg
Its speed before collision=2.5 m/s
Its speed after collision=2 m/s
Mass of second ball=0.6 kg
Momentum of 1st ball=mass of the ball*velocity
=0.3kg*2.5m/s
=0.75 kg m/s
Momentum of 2nd ball=mass of the ball*velocity
=0.6 kg*velocity of 2nd ball
Since the first ball undergoes head on collision with the second ball,
momentum of first ball=momentum of second ball
0.75 kg m/s=0.6 kg*velocity of 2nd ball
Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg
=1.25 m/s
Explanation:
Time = Distance/Speed
= 96km / (30km/h)
= 3.2h. (Or 3 hours and 12 minutes)
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
