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777dan777 [17]
3 years ago
13

HEYYYYYYYYYYYY ONCE AGAIN THIS IS ONLY 3 OF 30 PlS HELPPP

Mathematics
1 answer:
irakobra [83]3 years ago
8 0

Answer:

the 3rd one im pretty sure

Step-by-step explanation:

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The area of cross section of the large piston in a hydraulic machine is 1m² and that of small piston is 0.5m².calculate the forc
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Answer:

The force that must be applied on the small piston has a magnitude of 500 newtons.

Step-by-step explanation:

By Pascal's Principle, we know that pressure within any point of a closed hydraulic system is the same. Hence, the hydraulic machine can be described by the following model:

\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} (1)

Where:

F_{1}, F_{2} - Forces applied on the small and large pistons, in newtons.

A_{1}, A_{2} - Areas of the small and large pistons, in square meters.

If we know that A_{1} = 0.5\,m^{2}, A_{2} = 1\,m^{2} and F_{2} = 1000\,N, then the force to be applied on the small piston is:

F_{1} = \left(\frac{A_{1}}{A_{2}} \right)\cdot F_{2}

F_{1} = \left(\frac{0.5\,m^{2}}{1\,m^{2}} \right)\cdot (1000\,N)

F_{1} = 500\,N

The force that must be applied on the small piston has a magnitude of 500 newtons.

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