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.
A scientist would write that number as 1.49 x 10⁸ kilometers .
(Or, if the scientist is in France or the UK, he might write it as 1.49 x 10⁸ kilometres .)
Answer:
a) v₀ = 32.64 m / s
, b) x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
x = v₀ₓ t
y = 
t - ½ g t²
We look for the components of speed with trigonometry
sin 43 = v_{oy} / v₀
cos 43 = v₀ₓ / v₀
v_{oy} = v₀ sin 43
v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
t = x / v₀ cos 43
y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
y = x tan 43 - ½ g x² / v₀² cos² 43
1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
v₀ = √ (35280 / 33.11)
v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
y = t - ½ g t²
.y = v₀ sin43 t - ½ g t²
25 = 32.64 sin 43 t - ½ 9.8 t²
4.9 t² - 22.26 t + 25 = 0
t² - 4.54 t + 5.10 = 0
We solve the second degree equation
t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
t = (4.54 ± 0.46) / 2
t₁ = 2.50 s
t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
x = v₀ₓ t
x = v₀ cos 43 t
x = 32.64 cos 43 2.50
x = 59.68 m
Answer:
3k mph
Explanation:
don't take my answer it is wrong
