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3 years ago

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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of t

he 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:

sleet_krkn [62]3 years ago

6 0

Answer:

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.

fgiga [73]3 years ago

5 0

Answer:

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.

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