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Leno4ka [110]
2 years ago
5

A uniform cube with mass 0.700 kg and volume 0.0270 m3 is sitting on the floor. A uniform sphere with radius 0.400 m and mass 0.

800 kg sits on top of the cube.
1) How far is the center of mass of the two-object system above the floor? Express your answer in hcm unit.
Physics
1 answer:
Sav [38]2 years ago
6 0

Answer:

  44 1/3 cm

Explanation:

The cube has an edge length of ∛0.027 m = 0.3 m, so a center of mass (CoM) 15 cm above the floor.

The sphere's center of mass is 40 cm above the top of the cube, so is 70 cm above the floor. The weighted average of the CoM locations is ...

  ((15 cm)(0.700 kg) +(70 cm)(0.800 kg))/(0.700 kg +0.800 kg)

  = (10.5 kg·cm +56 kg·cm)/(1.500 kg) = 44.333... cm

The center of mass of the two-object system is 44 1/3 cm above the floor.

_____

<em>Comment on the units</em>

We're not familiar with "hcm" as a unit. We presume that you can convert the given answer to the units you desire.

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<u>Answer:</u>

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<u>Explanation:</u>

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  Acceleration = (40 i - 20 k)10^3m/s^2

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 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  Substituting

       s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m

     So final position = ((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m

                              =(708.15 i - 444.1 j + 200 k)*10^3km

    The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

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