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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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Answer:

a = \frac{2}{3}g

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As the disc is unrolling from the thread then at any moment of the time

We have force equation as

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7 0
2 years ago
Please help! will give brainlest!!!!!!!!!!!!
eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

\sum F = ma

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\sum F is the net force

m is the mass

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The bag is moving at constant speed, so its acceleration is zero:

a=0

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\sum F = 0

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

\sum F = F cos \theta - F_f = 0

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F cos \theta is the horizontal component of the applied force, with

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\theta=32.0^{\circ}

F_f is the force of friction

And solving for F_f, we find

F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N

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2 years ago
A dolphin swims due east for 1.90 km, then swims 7.20 km in the direction south of west. What are the magnitude and direction of
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Answer:

magnitude = 7.446 km, direction = 75.22° north of east

Explanation:

From the questions,

To get the the magnitude of the resultant vector we use Pythagoras theorem

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From the diagram,

y² = 1.9²+7.2²

y² = 55.45

y = √(55.45)

y = 7.446 km.

The direction of the dolphin is given as,

θ = tan⁻¹(7.2/1.9)

θ = tan⁻¹(3.7895)

θ = 75.22° north of east

Hence the magnitude of the resultant vector = 7.446 km, and it direction is 75.22° north of east

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The radius of the circular path is 1.5 m.

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