Question

A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

Answer 1

Answer:

The extra force exerted against the bottom is 5880 N.

Explanation:

Given that,

Diameter of the bottle= 2.00 cm

Diameter of the jug = 14.0 cm

Force = 120 N

We need to calculate the extra force exerted against the bottom

Using formula of force

$F_{2}=P\times A_{2}$

$F_{2}=\dfrac{F_{1}}{A_{1}}\times A_{2}$

$F_{2}=\dfrac{F_{1}}{\pi r^2}\times\pi r^2$

$F_{2}=\dfrac{F_{1}}{\pi\times\dfrac{d_{1}^2}{4}}\times\pi\times\dfrac{d_{2}^2}{4}$

$F_{2}=(\dfrac{d_{2}}{d_{1}})^2\times F_{1}$

Put the value into the formula

$F_{2}=(\dfrac{14.0}{2.00})^2\times120$

$F_{2}=5880\ N$

Hence, The extra force exerted against the bottom is 5880 N.