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2 years ago

11

A factory hires a consultant to recommend ways to improve its productivity. The consultant notices that the production floor is

strewn with tools and machines that are not continually in use. Which of the 5S philosophies should the consultant recommend to the management to establish better productivity?
A.
set in place
B.
sort
C.
standardize
D.
shine
E.
sustain

1 answer:

Romashka [77]2 years ago

8 0

Answer:

B

Explanation:

Keep only what is used. Remove all unnecessary items. Red tag these items for review.

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The question belongs to Electrical Engineering (Linear System).

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2 years ago

Write short notes on: (any four) a) Suspended ground floor b) Soil exploration c) Baulking of sand d) Bearing capacity of soil e

vredina [299]

Answer:

a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.

b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods

c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.

d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.

Explanation:

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6 0

3 years ago

For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea

miskamm [114]

Answer:

$k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

where $r_1$ and $r_2$ be the inner radius, outer radius of the annalus.

Explanation:

Let $r_1$ , $r_2$ and $L$ be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, $T_1=30^{\circ}C\\$ and at the outer surface, $T_2=40^{\circ}C$ .

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

$40 W/m^2$ .

$\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40$

$\Rightarrow q=2.4\pi L \;W$

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

$q=-kA\frac{dT}{dr}\;\cdots(i)$

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, $A=2\pi rL\;\cdots(ii)$

Variation of temperature w.r.t the radius of the annalus is

$\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}$

$\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)$

Putting the values from the equations (ii) and (iii) in the equation (i), we have

$q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}$

$\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}$

$\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)}$ [as $q=2.4\pi L$ , and $T_2-T_1=10 ^{\circ}C$ ]

$\Rightarrow k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0

3 years ago

Which of these include some basic characteristics of taxi services? (Select all that apply.)Transactions are often negotiated at

Nezavi [6.7K]

Answer:j

Explanation:

6 0

2 years ago

What is the pressure at the bottom of a 32m high water tower in gage pressure and absolute pressure (answers in kPa, and bar)

creativ13 [48]

Answer:

$P_{g}=313.6kPa, P_{g}=3.13bar, P_{a}=414.93kPa, P_{a}=4.15bar$

Explanation:

The gauge pressure is the one made by the water over the bottom of the tank without taking into account the pressure of the atmosphere. It can be calculated by using the following equation:

$P_{g}=\gamma _{H_{2}O} h$

in this case:

$\gamma _{H_{2}O}=9.8kN/m^{3}$

and h=32m

so the equation would then solve to:

$P_{g}=(9.8kN(m^{3})(32m)$

Which yields

$P_{g}=313.6kPa$

now, in order to find the bars, we must remember that:

1bar=100kPa, so

$313.6kPa*\frac{1bar}{100kPa}=3.136bar$

Now, in order to find the absolute pressure we need to add the atmospheric pressur to the pressures previously found.

$P_{a}=313.6kPa+101.325kPa$

$P_{a}=414.93kPa$

and likewise we can do the conversion of kPa to bars, so we get:

$414.93kPa*\frac{1bar}{100kPa}=4.15bar$

5 0

3 years ago