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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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Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para
Anastasy [175]

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

7 0
2 years ago
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0
2 years ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light w
Artemon [7]

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

7 0
2 years ago
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i
andrey2020 [161]

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field U=4\times 10^{-6}J

Value of inductance L =0.060 H

Energy stored in magnetic field is given by U=\frac{1}{2}Li^2

i=\sqrt{\frac{2U}{L}}

i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A

So the current flowing through rectangular toroid will be 0.01154 A

3 0
2 years ago
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