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2 years ago

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

Physics

1 answer:

slamgirl [31]2 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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Answer:

TRUE.

Explanation:

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The tape in a videotape cassette has a total

Softa [21]

Answer:

1.37 rad/s

Explanation:

Given:

Total length of the tape is, $d= 297$ m

Total time of run is, $t = 2.1$ hours

We know, 1 hour = 3600 s

So, 2.1 hours = 2.1 × 3600 = 7560 s

So, total time of run is, $t= 7560$ s

Inner radius is, $r = 10\ mm = 0.01\ m
$

Outer radius is, $R = 47\ mm = 0.047\ m
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Now, linear speed of the tape is, $v =\frac{d}{t}=\frac{297}{7560}=0.039\ m/s
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Let the same angular speed be $\omega$ .

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So, average radius is, $R_{avg}=\frac{R+r}{2}=\frac{0.047+0.01}{2}=\frac{0.057}{2}=0.0285\ m
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Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,

$\omega=\dfrac{v}{R_{avg}}\\\\\\\omega=\dfrac{0.039}{0.0285}\\\\\\\omega=1.37\ rad/s
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Therefore, the common angular speed of the reels is 1.37 rad/s.

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3 years ago

The electric motor of a model train accelerates the train from rest to 0.540 m/s in 27.0 ms. The total mass of the train is 610

Gnesinka [82]

Answer:

The value is $P =3.294 \ W$

Explanation:

From the question we are that

The velocity $v = 0.540 \ m/s$

The time taken is $t = 27.0 ms = 27.0 *10^{-3} \ s$

The total mass of the train is $m = 610 \ g = 0.610 \ kg$

Generally the average power delivered is mathematically represented as

$P =\frac{KE }{t}$

$P =\frac{ \frac{1}{2} * m * v^2 }{t}$

=> $P =\frac{ \frac{1}{2} * 0.610 * 0.540 ^2 }{ 27.0 *10^{-3}}$

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A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot

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To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

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at the bottom of the hill:
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