How large is the area of water found on Mars?

Physics

2 answers:

ehidna [41]3 years ago

8 0

More than five million cubic kilometers of ice have been identified.

Jlenok [28]3 years ago

7 0

Hello there!

More than five million cubic kilometers of ice have been identified at or near the surface of modern Mars, enough to cover the whole planet to a depth of 35 meters (115 ft).

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A planet's distance from the sun is 2.0x10^11 m. what is the orbital period?

Korolek [52]

Answer:

The square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit.

Explanation:

hope this helps.

4 0

2 years ago

Gold has a density of 19.3 g/cm3. What is the mass of a 5 cm3 block of gold?

Tems11 [23]

Do not forget that mass = volume x density
Mass of 1 cm^3 = Density[/tex]
$mass of 2 cm^3 = 19.3 g + 19.3 g = 2*19.3 g$
Then eventually we can find mass of 5 cm^3 : =
 $19.3 g + 19.3 g+19.3g+19.3g+19.3g= 5*19.3 g$
So the answer is D
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5 0

3 years ago

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

ohaa [14]

The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
$F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\$
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. $\alpha$ is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
$g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)
$
Our final equation is:
$g=g_0-g_0(a-1)cos^2(\alpha)$
Colatitude is:
$\alpha_c=90^\circ-\alpha$
The answer is:
$g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)$