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Mice21 [21]
3 years ago
7

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

40 cm2, how much weight can the large piston support?
Physics
1 answer:
weqwewe [10]3 years ago
6 0

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A   (N/m²)

P1=P2

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

F_{2} = \frac{F_{1}*A_{2}  }{A_{1}}  Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight  (W₂) can the large piston support

We replace data in the equation (1)

F_{2} = \frac{(500)*(40) }{2}

F₂ = 10000 N

W₂= F₂= 10000 N

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Answer:

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The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

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dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

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3 years ago
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Water does expand with heat (and contract with cooling), but the amount of expansion is pretty small. So when you boil a can filled with water and seal it, the water will contract slightly as it cools. The can may kink slightly, but that will be it. Actually, most likely the only things you will be able to see is then top and bottom will be sucked in and go concave. Just like a commercial can of beans.

Now if you have a can with a little water and a big air space, things are completely different.

As the water boils, water vapour is given off. Steam. Let it boils for a minute just to make sure (nearly) all the air is expelled and the can is filled with steam.

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the final temperature of the tea is 7.39⁰C.

Explanation:

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specific heat capacity of the tea, C = 4.184 JJ/g°C

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the final temperature of the tea, t₂ = ?

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The energy lost by the refrigerator is given by the following formula;

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-137,460 = 1569(t₂ - 95°C)

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Surface tension increases as intermolecular forces increases.

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