Ahost 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.

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

=

therefore

F2 =  × A2

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

A1 = π× = π× = 0.00042

A2 = π× = π× =  0.013

now substituting all values into F2 =  × A2

F2 =  × 0.013

F2 = 3.88 N

it is called kinetic energy

resonant frequency