Ask Question
4 May, 08:09

In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.400 m (1.31 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 12.00 millibar. The atmospheric pressure outside is 940 millibar. Calculate the force required to pull the two hemispheres apart.

+2
Answers (2)
  1. 4 May, 10:34
    0
    Force required to pull the two hemisphere = 46622.72N

    Explanation:

    Complete question (Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013*105 N/m2 ]

    The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.

    Pressure difference = (940 - 12) = 928 millibars.

    (928 x 100) = 92,800N/m^2.

    (92800 x 0.5024) = 46622.72N. force required to part the hemispheres.
  2. 4 May, 11:02
    0
    42 for each side (84)

    Explanation:

    [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013*105 N/m2 ]

    The contact area between the hemispheres is:-

    π x 0.440^2 = 0.6083m².

    The pressure difference is = (940 - 12) = 928 millibars.

    (928 x 100) = 92,800N/m².

    (92,800 x 0.6083) = 56,450.24N.

    Force required to part the hemispheres.

    (56,450.24/1,330) = 42 horses each side

    42 + 42 = 84 for each sides.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers