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emmainna [20.7K]
4 years ago
8

why is the meteor crater in arizona still visible when most impact craters created by meteorites hitting earth are not

Physics
2 answers:
damaskus [11]4 years ago
8 0
Good question. The Arizona impact location is very, very, unique. The crater was the Bigger than Cuba when it first struck. Due to sedimentary drop off its size is slowly decaying. It is currently about the size of 2-3 Carnival cruise ships.
joja [24]4 years ago
4 0

Answer:

The meteor crater in Arizona is still visible  due to the sedimentary nature of the materials present, mainly limestone, and the total absence of lava in the zone.

Most of the craters created by other meteorites that have collided with the earth are not visible at present because they were smaller and also impacted in areas where the soil materials were more compact and firm.

Explanation:

It is estimated that the impact produced by the crater 50,000 years ago, and was caused by an nickel-iron meteorite about 50 m long traveling at an approximate speed of 12 km/sec.  The impact created a temperature so extraordinarily high that it vaporized or melted all the rocks in the surrounding crust.

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Explanation:

It is given that, a solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ω.The sphere is slowly heated until it reaches its melting temperature, at which point it flattens into a uniform disk of thickness D/2.

The angular momentum remains conserved in this case. The relation between the angular momentum and the angular velocity is given by :

I_s\times \omega_s=I_d\times \omega_d

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I_s\ and\ I_d are moment of inertia of sphere and the disk respectively

Here, volume before = volume after

\dfrac{4}{3}\pi (D/2)^3=\pi r^2\times D/2

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L_i=\dfrac{2m(D/2)^2}{5}\times \omega_s

L_i=\dfrac{2mD^2}{20}\times \omega_s..........(1)

Final angular momentum,

L_d=I_f\times \omega_d

L_d=\dfrac{2mr^2}{5}\times \omega_d

L_d=\dfrac{2m(0.577D)^2}{5}\times \omega_d............(2)

From equation (1) and (2) :

\dfrac{2mD^2}{20}\times \omega_s=\dfrac{2m(0.577D)^2}{5}\times \omega_d

\dfrac{\omega_d}{\omega_s}=0.75

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