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GenaCL600 [577]
3 years ago
9

MGS and later orbiters found spectral evidence for what minerals on the Martian surface? Check all that apply.

Physics
1 answer:
slamgirl [31]3 years ago
6 0

Mars Global Surveyors (MGS) and later orbiters found the following  minerals on the  Martian surface;

  • Phyllosilicates
  • 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

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An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
jenyasd209 [6]

Answer:

a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

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he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug
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Answer:

207.4 N

Explanation:

The torque \tau  on a body is

\tau = r* F  where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

\tau = rF\sin \theta

Where \theta is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation \tau = rF\sin \theta

\tau = LF\sin \theta

Where L is the length of the wrench.

Making F the subject

F = \frac{\tau }{{L\sin \theta }}

Force required to pull the wrench is given as,

F = \frac{\tau }{{L\sin \theta }}

Substitute 47{\rm{ N}} \cdot {\rm{m}}  for \tau, 25 cm for L, and 115o for \theta  

\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}  

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When the starter motor on a car is engaged, there is a 320 A current in the wires between the battery and the motor. Suppose the
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Answer:

0.5

Explanation:

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