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
[tex]F_{2}=P\times A_{2}[/tex]
[tex]F_{2}=\dfrac{F_{1}}{A_{1}}\times A_{2}[/tex]
[tex]F_{2}=\dfrac{F_{1}}{\pi r^2}\times\pi r^2[/tex]
[tex]F_{2}=\dfrac{F_{1}}{\pi\times\dfrac{d_{1}^2}{4}}\times\pi\times\dfrac{d_{2}^2}{4}[/tex]
[tex]F_{2}=(\dfrac{d_{2}}{d_{1}})^2\times F_{1}[/tex]
Put the value into the formula
[tex]F_{2}=(\dfrac{14.0}{2.00})^2\times120[/tex]
[tex]F_{2}=5880\ N[/tex]
Hence, The extra force exerted against the bottom is 5880 N.
