A host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.30 cm diameter into the bottle, placing it in direct contact with the wine. The host is amazed when he pounds the cork into place and the bottom of the jug (with a 13.0 cm diameter) breaks away. Calculate the extra force (in N) exerted against the bottom if he pounded the cork with a 120 N force.

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usmanbiu usmanbiu · Answer 1 · 2019-11-09T21:26:50+01:00

Answer: 3.88 N

Explanation:

From the question we have the following:

cork diameter (D1) = 2.3 cm = 0.023 m

bottom diameter (D2) = 13 cm = 0.13 m

force on the cork (F1) = 120 N

force at the bottom (F2) = ?

we can get the force at the bottom by applying the formula below

[tex]\frac{F1}{A1}[/tex] = [tex]\frac{F2}{A2}[/tex]

therefore

F2 = [tex]\frac{F1}{A1}[/tex] × A2

where A1 and A2 are the areas of the cork and bottom respectively

A1 = π×[tex]r^{2}[/tex] = π×[tex](0.023 ÷ 2 )^{2}[/tex] = 0.00042

A2 = π×[tex]r^{2}[/tex] = π×[tex](0.13 ÷ 2 )^{2}[/tex] = 0.013

now substituting all values into F2 = [tex]\frac{F1}{A1}[/tex] × A2

F2 = [tex]\frac{120}{0.00042}[/tex] × 0.013

F2 = 3.88 N