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nevsk [136]
3 years ago
12

Three blocks are shown: Three blocks are shown. Block A has mass 3 kilograms, length 6 centimeters, height 4 centimeters, and wi

dth 2 centimeters. Block B has mass 1 kilograms, length 6 centimeters, height 2 centimeters, and width 4 centimeters. Block C has mass 2 kilograms, length 2 centimeters, height 6 centimeters, and width 4 centimeters. Which statement is correct? Block A has the greatest density. Block B has the greatest density. The density of Block A is equal to the density of Block B. The density of Block B is equal to the density of Block C.
Physics
1 answer:
Yakvenalex [24]3 years ago
5 0

Answer:

Block A has the greatest density.

Explaination:

Block A density:0.0625 kg/cm3

Block B density:0.020833 kg/cm3

Block C density:0.041667 kg/cm3

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Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

v = \frac{2 \pi r}{T}  

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T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

T^{2} = r^{3}

T = \sqrt{(133370 Km)^{3}}

T = \sqrt{(2.372x10^{15} Km)}

T = 4.870x10^{7} Km

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( 1.50x10^{8} Km )

1 AU is defined as the distance between the earth and the sun.

\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU

T = 0.324 AU

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

T = \frac{0.324 AU}{1 AU} . 1 year

T = 0.324 year

That can be expressed in units of days

T = \frac{0.324 year}{1 year} . 365.25 days  

T = 118.60 days

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v = \frac{2 \pi r}{T}

v = \frac{2 \pi (133370 Km)}{(118.60 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

118.60 days .\frac{86400 s}{1 day} ⇒ 10247040 s

133370 Km .\frac{1000 m}{1 Km} ⇒ 133370000 m

v = \frac{2 \pi (133370000 m)}{(10247040 s)}

v = 81.778 m/s

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

T^{2} = r^{3}

T = \sqrt{(69000 Km)^{3}}

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v = \frac{2 \pi r}{T}

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v = 114.483 m/s

 

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