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Burka [1]
2 years ago
8

Pistons are fitted to two cylindrical chambers connected through a horizontal tube to form a hydraulic system. The piston chambe

rs and the connecting tube are filled with an incompressible fluid. The cross-sectional areas of piston 1 and piston 2 are A1 and A2, respectively. A force F1 is exerted on piston 1. Rank the resultant force F2 on piston 2 that results from the combinations of F1, A1, and A2 given from greatest to smallest. If any of the combinations yield the same force, give them the same ranking. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)
F1 = 4.0 N; A1 = 0.9 m2; and A2 = 1.8 m2
F1 = 2.0 N; A1 = 0.9 m2; and A2 = 0.45 m2
F1 = 2.0 N; A1 = 1.8 m2; and A2 = 3.6 m2
F1 = 4.0 N; A1 = 0.45 m2; and A2 = 1.8 m2
F1 = 4.0 N; A1 = 0.45 m2; and A2 = 0.9 m2
F1 = 2.0 N; A1 = 1.8 m2; and A2 = 0.9 m2
Physics
1 answer:
Ivenika [448]2 years ago
6 0

Answer:

order   d> a = e> c> b = f

Explanation:

Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form

      P₁ = P₂

Using the definition of pressure

      F₁ / A₁ = F₂ / A₂

      F₂ = A₂ /A₁   F₁

Now we can examine the results

a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2

     F₂ = 1.8 / 0.9 4

     F₂a = 8 N

b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2

    F₂b = 0.45 / 0.9 2

    F₂b = 1 N

c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2

    F₂c = 3.6 / 1.8 2

    F₂c = 4 N

d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2

    F₂d = 1.8 / 0.45 4.0

    F₂d = 16 m2

e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2

   F₂e = 0.9 / 0.45 4

   F₂e = 8 N

f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2

   F₂f = 0.9 / 1.8 2.0

   F₂f = 1 N

Let's classify the structure from highest to lowest

F₂d> F₂a = F₂e> F₂c> F₂b = F₂f

I mean the combinations are

 d> a = e> c> b = f

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Suppose the truck that’s transporting the box In Example 6.10 (p. 150) is driving at a constant speed and then brakes and slows
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Answer:

Friction acts in the opposite direction to the motion of the truck and box.

Explanation:

Let's first review the problem.

A moving truck applies the brakes, and a box on it does not slip.

Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.

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5 0
1 year ago
A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i
sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, v_{Ci}=25\ m/s

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, v_{Ti}=20\ m/s

After the collision, velocity of the car, v_{Cf}=18\ m/s

Let v is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

initial\ momentum=final\ momentum

1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v

210000-21600=9000\ kg\times v

v=20.93\ m/s

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0
1 year ago
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