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3 years ago

What 2 forces move the secondary piston ahead?

Engineering

1 answer:

jekas [21]3 years ago

5 0

Answer:

The primary piston activates one of the two subsystems. The hydraulic pressure created, and the force of the primary piston spring, moves the secondary piston forward.

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In fully developed laminar flow in a circular pipe the velocity at R/2 (mid-way between the wall surface and the centerline) is

Butoxors [25]

Answer:

$u_{max} = 17.334\,\frac{m}{s}$

Explanation:

Let consider that velocity profile inside the circular pipe is:

$u(r) = 2\cdot U_{avg} \cdot \left(1 - \frac{r^{2}}{R^{2}} \right)$

The average speed at $r = \frac{1}{2} \cdot R$ is:

$U_{avg} = \frac{13\,\frac{m}{s} }{2\cdot \left(1-\frac{1}{4} \right)}$

$U_{avg} = 8.667\,\frac{m}{s}$

The velocity at the center of the pipe is:

$u_{max} = 2\cdot U_{avg}$

$u_{max} = 17.334\,\frac{m}{s}$

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3 years ago

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8. What is the purpose of the 300 Log?

Gnom [1K]

Answer:

“An OSHA 300 log is where companies record the injuries that occur at the workplace,” said Luna. “By law, they have to report all the injuries to OSHA.” The OSHA law gives workers and their unions the right to have access to injury logs.

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3 years ago

Why it is important to prepare first the materials and tools carpentry before doing the tasks?

professor190 [17]

Answer:

It is important to inspect and check materials and tools for defects and damage before receiving them so that you can ask for replacements for those that you found .

Explanation:

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3 years ago

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The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu

marishachu [46]

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

3 0

4 years ago

what are the characteristics of an ideal fluid the general relation between shear stress and velocity gradient

Dafna11 [192]

Answer:

ideal fluid follow Newtonian law

that is, shear stress is directly proportional to rate change of shear strain.

watch handwritten explanation

6 0

3 years ago