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3 years ago

6

El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es

de 32 cm de diámetro. ¿Qué presión de aire se tendrá de aire se tendrá que usar para levantar un automóvil de 17,640 N?`

1 answer:

topjm [15]3 years ago

3 0

Answer:

P₁ = 219.3 Pa

Explanation:

This fluid mechanics problem, we can use that the pressure is distributed with the same value throughout the system, which is Pascal's principle.

Let's use the subinidce1 for the small diameter and the subscript 2 for the larger diameter.

P₁ = P₂

pressure is defined by

P = F / A

we subtitute

F₁ / A₁ = F₂ / A₂

F₁ = F₂ A₁ / A₂

the area in a circle is

A = π r² = π d² / 4

we substitute

F₁ = F₂ (d₁ / d₂)²

we calculate

F₁ = 17640 (2/32)²

F₁ = 68.9 N

Having the force to be applied we can find the air pressure on the small plunger

P₁ = F₁ / A₁

P₁ = F₁ 4 / π d₁²

let's calculate

P₁ = 68.9 4 / (π 0.02²)

P₁ = 219.3 Pa

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A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

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Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago

Lubov Fominskaja [6]

question: Please help!!!

If a bottle is being squeezed with a force of 10 Newtons and the area of the bottle is 15

squared inches. What is the pressure??

Answer:

1025.64 N/m²

Explanation:

Pressure: This can be defined as the ratio of force to area. The si unit of pressure is N/m².

From the question,

P = F/A........................ Equation 1

Where F = Force, A = Area.

Given: F = 10 Newtons, A = 15 Squared Inches = (15×0.00065) = 0.00975 m²

Substitute these values into equation 1

P = 10/0.00975

P = 1025.64 N/m²

Hence the pressure of the bottle is 1025.64 N/m²

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