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schepotkina [342]
2 years ago
7

if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice

of A1, what will F2 be in comparison to F1
Physics
1 answer:
Firlakuza [10]2 years ago
7 0

Answer:

2F_{1}

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}

F_{1}= \frac{F_{2}}{2}\\

F_{2}= 2F_{1}

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The compete question is shown on the first uploaded question

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From the question we are told that

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Generally the distance of the listener to the first speaker is mathematically represented as

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Generally the distance of the listener to second speaker at its new position is  

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=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

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