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lisov135 [29]
3 years ago
9

The surface of Mercury has many more craters than the surface of Earth has. Which of the following statements best explains why

Mercury has many more craters than Earth has?
Geography
2 answers:
IgorC [24]3 years ago
8 0

Answer:

Lacking the actual choices I assume mass, magnetic field, Atmosphere, or the Moon.

Explanation:

Mercury is 4,879 km, which is approximately 38% the diameter of Earth. ie barely larger than the moon.  answer was also supposed to include the presence of the moon, a magnetic field only 1.1 % the strength of Earth, and lacks a protective atmosphere to burn up/ reduce the size of meteors.

 

Neko [114]3 years ago
4 0

D Mercury has a thinner atmosphere than eart

Explanation:

Hope it helps

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. If the fastest passenger aircraft can fly 0.45 km/s (1000 mph), how long would it take to reach the sun? The galactic center?
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The distance to the sun from the Earth is approximately equal to 1.496\times 10^{8}\,km, if the passenger travels at constant speed, then the time needed to reach the sun is calculated by the following kinematic formula:

\Delta t = \frac{s}{v} (1)

Where:

s - Travelled distance, measured in kilometers.

v - Speed of the passenger aircraft, measured in kilometers per second.

\Delta t - Travelling time, measured in seconds.

If we know that s = 1.496\times 10^{8}\,km and v = 0.45\,\frac{km}{s}, then the travelling time is:

\Delta t = \frac{1.496\times 10^{8}\,km}{0.45\,\frac{km}{h} }

\Delta t = 3.324\times 10^{8}\,s

\Delta t = 3847.736\,days

\Delta t = 10.542\,years

The passenger aircraft would take 10.542 years to reach the Sun from the Earth.

The distance between the Earth and the galactic center is approximately equal to 2.460\times 10^{17}\,km. If the passenger travels at constant speed and if we know that s = 2.460\times 10^{17}\,km and v = 0.45\,\frac{km}{s} , then the travelling time is:

\Delta t = \frac{2.460\times 10^{17}\,km}{0.45\,\frac{km}{s} }

\Delta t = 5.467\times 10^{17}\,s

\Delta t = 6.327\times 10^{12}\,days

\Delta t = 1.733\times 10^{10}\,years

The passenger aircraft would take 1.733\times 10^{10} years to reach the gallactic center.

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3 years ago
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