Thank you for visiting A force of 3600 N is exerted on a piston that has an area of tex 0 030 text m 2 tex What force is. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Answer:
18,000 N
Explanation:
apex
Thank you for reading the article A force of 3600 N is exerted on a piston that has an area of tex 0 030 text m 2 tex What force is. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
- You are operating a recreational vessel less than 39 4 feet long on federally controlled waters Which of the following is a legal sound device
- Which step should a food worker complete to prevent cross contact when preparing and serving an allergen free meal A Clean and sanitize all surfaces
- For one month Siera calculated her hometown s average high temperature in degrees Fahrenheit She wants to convert that temperature from degrees Fahrenheit to degrees
Rewritten by : Jeany
For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
[tex]p_1 = p_2[/tex]
which becomes
[tex] \frac{F_1}{A_1}= \frac{F_2}{A_2} [/tex]
where
[tex]F_1=3600 N[/tex] is the force on the first piston
[tex]A_1=0.030 m^2[/tex] is the area of the first piston
[tex]F_2[/tex] is the force on the second piston
[tex]A_2=0.015 m^2[/tex] is the area of the second piston
If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
[tex]F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N[/tex]
[tex]p_1 = p_2[/tex]
which becomes
[tex] \frac{F_1}{A_1}= \frac{F_2}{A_2} [/tex]
where
[tex]F_1=3600 N[/tex] is the force on the first piston
[tex]A_1=0.030 m^2[/tex] is the area of the first piston
[tex]F_2[/tex] is the force on the second piston
[tex]A_2=0.015 m^2[/tex] is the area of the second piston
If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
[tex]F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N[/tex]