Intensive properties and extensive properties are types of physical properties of matter. The terms intensive and extensive were first described by physical chemist and physicist Richard C. Tolman in 1917. Here's a look at what intensive and extensive properties are, examples of them, and how to tell them apart.
Intensive Properties
Intensive properties are bulk properties, which means they do not depend on the amount of matter that is present. Examples of intensive properties include:
Boiling point
Density
State of matter
Color
Melting point
Odor
Temperature
Refractive Index
Luster
Hardness
Ductility
Malleability
Intensive properties can be used to help identify a sample because these characteristics do not depend on the amount of sample, nor do they change according to conditions.
Extensive Properties
Extensive properties do depend on the amount of matter that is present. An extensive property is considered additive for subsystems. Examples of extensive properties include:
Volume
Mass
Size
Weight
Length
The ratio between two extensive properties is an intensive property. For example, mass and volume are extensive properties, but their ratio (density) is an intensive property of matter.
While extensive properties are great for describing a sample, they aren't very helpful identifying it because they can change according to sample size or conditions.
Way to Tell Intensive and Extensive Properties Apart
One easy way to tell whether a physical property is intensive or extensive is to take two identical samples of a substance and put them together. If this doubles the property (e.g., twice the mass, twice as long), it's an extensive property. If the property is unchanged by altering the sample size, it's an intensive property.
Answer:
Step-by-step explanation:
1) x=14 y=40
2)a=10 b=37
3)u=62 v=59
4)s=9 t=14
5)u=66 v=38
hope this helps
To solve this, it might be easier to draw it out (see the picture below). I split it into two triangles and used trig functions to find the altitude. I used the big triangle to find theta, and the used theta to find the side of the altitude. *remember that sine= opposite/hypotenuse*
Solve for R:
R + 3 = -(1/2 + 6)
Put 1/2 + 6 over the common denominator 2. 1/2 + 6 = (2×6)/2 + 1/2:
R + 3 = -(2×6)/2 + 1/2
2×6 = 12:
R + 3 = -(12/2 + 1/2)
12/2 + 1/2 = (12 + 1)/2:
R + 3 = -(12 + 1)/2
12 + 1 = 13:
R + 3 = -13/2
Subtract 3 from both sides:
R + (3 - 3) = -13/2 - 3
3 - 3 = 0:
R = -13/2 - 3
Put -13/2 - 3 over the common denominator 2. -13/2 - 3 = (-13)/2 + (2 (-3))/2:
R = (-13)/2 - (3×2)/2
2 (-3) = -6:
R = (-6)/2 - 13/2
(-13)/2 - 6/2 = (-13 - 6)/2:
R = (-13 - 6)/2
-13 - 6 = -19:
Answer: R = (-19)/2
Solution:
Total number of Days that Pooja's plant survived= 2 + 98 = 100 Days=(t)
Longest Height obtained by plant = 30 cm= h
As the two , i.e number of Days lived by plant and it's longest height obtained are directly proportional.
We will solve it by unitary method.
100 Days = 30 cm
1 cm = Days
If height of the plant is h cm ,and amount of time taken is t days, then
h(t) cm = ,for t=0,1,2,3,4,5,6....we get different values of h.
