How are the surface of Mercury and Mars similar to the surface of the Earth?

Physics

1 answer:

scZoUnD [109]3 years ago

8 0

Answer:

The terrestrial planets all have rocky surfaces that feature mountains, plains, valleys and other formations. ... Mars has very low atmospheric pressure, and Mercury has almost none, so craters are more common on these planets.

Explanation:

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If you're ever standing on a mountaintop when a dark cloud passes overhead and your hair stands up, get off the mountain fast. H

OleMash [197]

Answer:

The hairs would have acquired charge by the passing of dry winds resulting in the loss of electron.

Explanation:

While standing on the top of a mountain if a person gets its hairs stand up after a cloud passes over, this might happen due to the static electric charges on the lower surface of the cloud are opposite in nature to that of hairs which the hairs would have acquired by the passing of dry winds which would have resulted in the loss of electron from the hair tip.

Similar case happens when we rub a dry plastic ruler or a dry plastic comb on our hairs.

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4 years ago

Which statement is true about the part of the electromagnetic spectrum that human eyes can detect?

LUCKY_DIMON [66]

Your answer is B. The human eye can only detect color and Tv waves

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3 years ago

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A circular test track for cars has a circumference of 3.5 km . A car travels around the track from the southernmost point to the

Marta_Voda [28]

<span>The car would have traveled exactly one-half of the circumference of the track, since it would have gone from one extreme point to its opposite extreme point. This would be equal to (3.5 / 2), or 1.75 km. The northernmost point would be 1.75km away from the southernmost point.</span>

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3 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$