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
Even though the wind "tries" to flow from high pressure to low pressure, the turning of the Earth causes the air flow to turn to the right (in the Northern Hemisphere), so the jet stream flows around the air masses, rather than directly from one to the other.
The answer would be 5.6x10^5
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
V(r) = ![-[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=-%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
