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irina1246 [14]
3 years ago
7

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capp

ed at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the smaller piston in order to start lifting the car
Physics
1 answer:
Bond [772]3 years ago
7 0

Answer:

= 925.92 N

≅ 926N

Explanation:

Pressure due to car = pressure due to applied force  

12000/18^2 = Force / 5^2

force = 12000 * 25/ 324

= 925.92 N

For equilibrium

Pressure1 = Pressure2

A1F1 = A2F2

12000*pi*(5^2) = F2 ( pi)*(18^2)

so, F2 = Applied force to lift car = 925.92 N

Pascal's principle

Pressure1 = Pressure2

F1/A1 = F2/A2 (F=force and A=area)

A1 =Pi*(0.05)²

A2 =Pi(0.18)²

F2=12000

F1 = 12000*(0.05)² / (0.18)² = 926N

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A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod
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(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
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