Answer:
Intensive properties
Density
Color
temperature
Melting point
Extensive properties
Mass
Volume
Total Energy
Explanation:
Intensive properties: In Physics, Intensive properties which are not depend of the amount of matter in a sample, It only depends of the type of matter, some examples of intensive properties are:
1. Density: It is a intensive property. It can explain better with a example: the water density is 1000 kg/m3, So if we have 1 liter or 1000 liters of water the density will be the same for the two samples.
2. Color: Solid sodium chloride is white. If you have 2 samples the first recipient with 2 kilograms of NaCl and the second with 10 kilograms of NaCl. The color of the substance does not depend on the amount of the substance.
As was mentioned before the same theory is applied to temperature and melting point concepts.
On the other hand,
Extensive properties are properties of the matter which depend on the amount of matter that is present in the system or sample. some examples are:
1. Mass: It is a property that measures the amount of matter that an object contains. For example, 10 kilograms of solid Copper contains a higher mass than 2 kilograms of the same metal.
2. Volume: It is a property which measures the space occupied by an object or a substance. For example, the space occupied by a glass of milk is lower than the space occupied by a bottle of milk, Then the volume of the glass of milk is lower than the volume of the bottle of milk.
3. Finally the total energy is contained in molecules and atoms that constituted systems so, if the amount of matter increases the number of molecules too, then the total energy will increase.
I hope it helps you.
Answer:
2 kg
Explanation:
Acceleration = 5 m/s^2
Force = 10 N
Force = mass * acceleration
mass = force / acceleration
mass = 10 / 5
mass = 2 kg
Answer:
53.13 °
Explanation:
In order to do this, we just need to apply the following:
tanα = Dy/Dx
Where:
Vy: speed of the ball in the y axis.
Vx: speed of the ball in the x axis.
At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.
According to this, then:
tanα = (40/30)
tanα = 1.3333
α = tan⁻¹(1.3333)
<h2>
α = 53.13°</h2>
This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.
Hope this helps
