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alukav5142 [94]
2 years ago
8

Which apparatus is likely to carry a ladder? (There may be more than one answer.)

Engineering
1 answer:
Aloiza [94]2 years ago
4 0
B and D
hope this helped
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Give me source code of Simple openGL project. ( without 3D or Animation) simple just.
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Use GitHub or stackoverflow for this answer

Explanation:

It helps with programming a lot

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Water enters a boiler at a temperature of 110°F. The boiler is to produce 2000 lb/hr of steam at a pressure of 130 psia. How man
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A 50-kN hydraulic press performs pressing and clamping actions. The clamping cylinder force is 4 kN. The pressing cylinder strok
galben [10]

Answer:

The attached figure shows the hydraulic circuit using one sequence valve to control two simultaneous operations performed in proper sequence in one direction only. In the other direction, both the operations are simultaneous.

When we keep the 4/2 DCV in crossed arrow position, oil under pressure is supplied to the inlet port of the sequence valve. It directly flows to Head end port-1. Hence Cylinder 'C1' extends first.

By the end of the extension of cylinder 'C1', pressure in the line increases and hence poppet of sequence valve is lifted off from its seat and allows oil to flow to port-2 and hence, Cylinder 'C2 extends completing the pressing operation.

In the straight-arrow position of 4/2 DCV the oil under pressure reaches the rod end of both the cylinders C1 and C2 simultaneously through port-3. This causes both the cylinders to retract simultaneously.

Also, a Flow control valve is provided tho control the velocity of clamping

Explanation:

find attached the figure

4 0
3 years ago
A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the
muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, \L_e = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

P =\frac{\pi^2EI}{L_e^2}

where, I is the moment of inertia

on substituting the respective values, we get

10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}

or

I = 13.13 in⁴

also for circular cross-section

I = \frac{\pi}{64}\times d^4

thus,

13.13 = \frac{\pi}{64}\times d^4

or

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0
3 years ago
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