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
Answer:
D: The distance between the particles decreases
Explanation:
Taking away energy slows down molecules, like how you slow down when you are cold (I think)
When you say full valence shell, are you talking about a valence electron shell?
I am learning about atoms and i know a little bit
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:

where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:

where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:

- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:

- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

- Replacing by the givens in (5), we can solve for μsmín, as follows:
