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3 years ago

13

In a hydraulic device, the surface area of the small piston is 1 cm2 and the surface area of the large piston is 8 cm2. If 10 N

of force is put on the small piston, how much force will be exerted on the large piston.

1 answer:

kotykmax [81]3 years ago

7 0

Hence the force is 80 N

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Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

<em>by using conservation momentum, which stated that the initial momentum is equal to the final momentum.</em>

<em /> $m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em>so since the bottle is at rest firstly, therefore </em> $v_{i2} =0$ <em />

<em /> $m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em /> $m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ <em> </em><em>equation 1</em>

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

<em /> $m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$ <em />

<em>collect the like terms</em>

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

<em />

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