Ask question

Ask question

All categories

3 years ago

11

Two hydraulic cylinders maintain a pressure of 1200 kPa. One has a cross sectional area of 0.01 mAA2EEAA the other 0.03 mAA2EEAA

. To deliver a work of 1 kJ to the piston how large a displacement (V) and piston motion H is needed for each cylinder? g

1 answer:

fredd [130]3 years ago

5 0

Answer:

Answered

Explanation:

W = ∫ F dx = ∫ P dV = ∫ PA dx = PA* H = P∆V

∆V = W/P

= 1 kJ / 1200 kPa = 0.0 833 m3

Both cases the height is H = ∆V/A

IN case 1 ( sectional area = 0.01 m^2)

$H_1 = \frac{0.000833}{0.01} = 0.0833$

in case 2 (sectional area = 0.03 m^2)

$H_2 = \frac{0.000833}{0.03} = 0.0278$

You might be interested in

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

7 0

3 years ago

Daily life examples of pressure of solids on solids?

Ipatiy [6.2K]

Explanation:

There's a massive amount, just think of anything everyday. Like a table on the floor, or when your walking around and putting pressure on the floor. When you squeeze something which is solid. Anything like that will do.

6 0

3 years ago

The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the

8090 [49]

Compute first for the vertical motion, the formula is:

y = gt²/2

0.810 m = (9.81 m/s²)(t)²/2

t = 0.4064 s

whereas the horizontal motion is computed by:

x = (vx)t

4.65 m = (vx)(0.4064 s)

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed.

(vy) = gt

(vy) = (9.81 m/s²)(0.4064 s)

(vy) = 3.99 m/s

speed with which it hit the ground:

v = sqrt[(vx)² + (vy)²]

v = sqrt[(11.44 m/s)² + (3.99 m/s)²]

v = 12.12 m / s

6 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t

Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

$Ff = m*a$ where Ff = μ * N and $a = \frac{V^2}{r}=\omega^2*r$

N - m*g = 0 So, N = m*g. Replacing everything on the original equation:

$\mu*m*g = m*\omega^2*r$ (eq2)

Solving for r:

$r = \frac{\mu*g}{\omega^2}=1.41m$

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0

3 years ago