Mars Global Surveyors (MGS) and later orbiters found the following minerals on the Martian surface;
- Carbonate
- Sulfates
- Iron oxide
The Mars Global Surveyors (MGS) and later orbiters suggest that the Martian crust contains a higher percentage of volatile elements such as Sulphur and chlorine than the Earth's crust does.
These scientists also conclude that the most abundant chemical elements in the Martian crust are those found in Igneous rock.
These elements include the following;
- Silicon,
- Oxygen,
- Iron,
- Magnesium,
- Aluminum,
- Calcium, and
- Potassium.
They also, suggest that hydrogen is found in ice (water) while carbon is found in carbon dioxide and carbonates.
From the given options the minerals found in Martian surface include;
- Phyllosilicates ------ these are sheet of silicate minerals
- Carbonate
- Sulfates
- iron oxide
Learn more here: brainly.com/question/20470323
i do believe the answer is c not sure but correct me if im wrong
Hey there!
Your answer: Spilling breaker
Spilling breaker usually occurs when a beach or ocean is flat, and as the waves of the wind continues to happen, slowly the region would eventually become a slope.
It's almost like play-dough. Let's say that we set a perfect flat surface of play-dough on the table. As we continue slide our hands one direction, doesn't the play dough have more on one side than the other? It eventually contains a slope if you add enough from the first place.
Your answer: Spilling breaker
Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
