Answer:
Its d
atome contain
negative electrons,
positive protons and uncharged neutrons.
Explanation:
Answer:
Solids
:A solid has a definite shape and volume because the molecules that make up the solid are packed closely together and move slowly. Solids are often crystalline; examples of crystalline solids include table salt, sugar, diamonds, and many other minerals. Solids are sometimes formed when liquids or gases are cooled; ice is an example of a cooled liquid which has become solid. Other examples of solids include wood, metal, and rock at room temperature. Liquids
: A liquid has a definite volume but takes the shape of its container. Examples of liquids include water and oil. Gases may liquefy when they cool, as is the case with water vapor. This occurs as the molecules in the gas slow down and lose energy. Solids may liquefy when they heat up; molten lava is an example of solid rock which has liquefied as a result of intense heat. Gases
: A gas has neither a definite volume nor a definite shape. Some gases can be seen and felt, while others are intangible for human beings. Examples of gases are air, oxygen, and helium. Earth's atmosphere is made up of gases including nitrogen, oxygen, and carbon dioxide. Plasma: Plasma has neither a definite volume nor a definite shape. Plasma often is seen in ionized gases, but it is distinct from a gas because it possesses unique properties. Free electrical charges (not bound to atoms or ions) cause the plasma to be electrically conductive. The plasma may be formed by heating and ionizing a gas. Examples of plasma include stars, lightning, fluorescent lights, and neon signs.
Explanation:
All spontaneous processes release free energy
Explanation:
Light rays that pass close to the black hole get caught and cannot escape. so what ends up happening to the light is that it starts bending due to the strong gravitational force of the black hole.
Refer to the figure shown below, which is based on the given figure.
d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.
Part A
From the geometry, obtain
d = X cos(α) (1a)
h = X sin(α) (1b)
The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α) (2a)
u = v₀ cos(θ - α) (2b)
If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0 (3a)
ut = d (3b)
Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain

![4.9[ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ]^{2} - v_{0} sin(\theta - \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ] - X sin \alpha = 0](https://tex.z-dn.net/?f=4.9%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%20%5D%5E%7B2%7D%20-%20v_%7B0%7D%20sin%28%5Ctheta%20-%20%20%5Calpha%20%29%20%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%5D%20-%20X%20sin%20%5Calpha%20%20%3D%200)
Hence obtain
![aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta - \alpha )}]^{2} \\ b = cos \alpha \, tan(\theta - \alpha ) + sin \alpha](https://tex.z-dn.net/?f=aX%5E%7B2%7D-bX%3D0%20%5C%5C%20where%20%5C%5C%20a%3D4.9%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%5D%5E%7B2%7D%20%5C%5C%20%20b%20%3D%20cos%20%5Calpha%20%5C%2C%20%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%20%2B%20sin%20%5Calpha%20)
The non-triial solution for X is

Answer:
![X= \frac{sin \alpha + cos \alpha \, tan(\theta - \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta - \alpha )} ]^{2}}](https://tex.z-dn.net/?f=X%3D%20%5Cfrac%7Bsin%20%5Calpha%20%20%2B%20cos%20%5Calpha%20%20%5C%2C%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%7B4.9%20%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20%5C%2C%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%20%20%5D%5E%7B2%7D%7D%20)
Part B
v₀ = 20 m/s
θ = 53°
α = 36°
sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423
X = 0.8351/(4.9*0.0423²) = 101.46 m
Answer: X = 101.5 m