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2 years ago

A retail business, owned by share holders and having centralized decision making for their multiple store locations is called:

Engineering

1 answer:

lara [203]2 years ago

4 0

Answer:

An enterprise that produces goods or provides services, usually to make a profit.

Explanation:

You might be interested in

What are 4 activities that you do in your daily life that DO require energy?

SashulF [63]

Answer: Some activities that I do in my daily life that require energy are:

1. Doing ballet

2. Studying

3. Walking up and down stairs

4. Stretching

5. Running on the treadmill

Hope this helps! :)

Explanation:

4 0

3 years ago

The modulus of elasticity for a ceramic material having 6.0 vol% porosity is 303 GPa. (a) Calculate the modulus of elasticity (i

Phantasy [73]

Answer:

modulus of elasticity for the nonporous material is 340.74 GPa

Explanation:

given data

porosity = 303 GPa

modulus of elasticity = 6.0

solution

we get here modulus of elasticity for the nonporous material Eo that is

E = Eo (1 - 1.9P + 0.9P²) ...............1

put here value and we get Eo

303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )

solve it we get

Eo = 340.74 GPa

8 0

3 years ago

Explain briefly the power analogy between the fluid “water” and the Word of God.

melomori [17]

Throughout the Word of God, water is a symbol, a "type," of the spirit of God, and is ... to represent holy spirit, that is, the divine nature and power of God. ... The average person can go about 60 days without food before he ... record illustrating the analogous relationship between water and the holy spirit.

3 0

3 years ago

A long rod of 60-mm diameter and thermophysical properties rho= 8000 kg/m3, c= 500 J/kg·K, and k= 50 W/m·K is initially at a uni

Dvinal [7]

Answer:

Tc = = 424.85 K

Explanation:

Data given:

D = 60 mm = 0.06 m

$\rho = 8000 kg/m^3$

k = 50 w/m . k

c = 500 j/kg.k

$h_{\infty} = 1000 w/m^2$

$t_{\infity} = 750 k$

$t_w = 500 K$

$surface area = As = \pi dL$

$\frac{As}{L} = \pi D = \pi \timeS 0.06$

HEAT FLOW Q is

$Q = h_{\infty} As (T_[\infty} - Tw)$

$= 1000 \pi\times 0.06 (750-500)$

= 47123.88 w per unit length of rod

volumetric heat rate

$q = \frac{Q}{LAs}$

$= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}$

$q = 1.66\times 10^{7} w/m^3$

$Tc = \frac{- qR^2}{4K} + Tw$

$= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 56} + 500$

= 424.85 K

7 0

3 years ago

Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compressio

SVETLANKA909090 [29]

This question is incomplete, the complete question is;

Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.

Use Fy=50 ksi and assume Cb=1.0 (if needed).

Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft

Explanation:

Given that;

section W 21 x 73 steel beam;

now from the steel table table for this section;

Zx = Sx = 151 in³

also given that; fy = 50 ksi and Cb = 1.0

QMn = 0.9 × Fy × Zx

so we substitute

QMn = 0.9 × 50 × 151

QMn = 6795 k-inch

we know that;

12inch equals 1 foot

QMn = 6795 k-inch / 12

QMn = 566.25 f-ft

Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft

7 0

3 years ago