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
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Answer:
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
Explanation:
Answer:A
Explanation:
It’s bigger I am not sure
Answer:
4.7 m³
Explanation:
We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2
* Givens :
P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K
( We must always convert the temperature unit to Kelvin "K")
* What we want to find :
V2 = ?
* Solution :
101 × 2 / 300.15 = 40 × V2 / 283.15
V2 × 40 / 283.15 ≈ 0.67
V2 = 0.67 × 283.15 / 40
V2 ≈ 4.7 m³
The calculated mutual inductance is 8.544 x 10⁻⁵ H.
Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.
Length of the solenoid= 5.0 cm
Area of cross-section=1.0 cm²
no of spaced turns=300 turns
turns of insulated wire=180 turns
Mutual inductance (M) = μ₀μr N1N2 A/ L
=(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5
=79.12 x 10⁻¹⁰ x 54000 / 5
=8.544 x 10⁻⁵ H
hence, the mutual inductance is 8.544 x 10⁻⁵ H.
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