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2 years ago

An HVAC contracting company sent out three work crews to complete three

Engineering

1 answer:

Doss [256]2 years ago

6 0

Answer: 156 hours

Explanation:

Crew One worked 10-, 9-, 11-, 12-, and 9-hours for the week. The total hours worked by Crew One will be:

= 10 + 9 + 11 + 12 + 9

= 51 hours.

Crew Two worked 9, 12, 12, 9 and 9 hours. The total hours worked by Crew Two for the week will be:

= 9 + 12 + 12 + 9 + 9

= 51 hours

Crew Three worked 12-, 12-, 10-, 9-, and 11-hour daily. The total hours worked by Crew Three for the week will be:

= 12 + 12 + 10 + 9 + 11

= 54 hours

The total hours worked by the three crews for the week will be:

= 51 hours + 51 hours + 54 hours

= 156 hours

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An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formatio

saveliy_v [14]

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

$K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}$

Put the value into the formula

$K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}$

$K_{H}=41.9\ m/d$

We need to calculate the vertical conductivity

Using formula of vertical conductivity

$K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}$

Put the value into the formula

$K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}$

$K_{V}=37.2\ m/d$

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

3 0

2 years ago

Where can you find free air pods that look real

Tema [17]

You can find air pods that look real on letgo. or you can go to wish.com but if you want a good pair jus get the real ones

7 0

3 years ago

The maximum stress that a bar will withstand before failing is called • Rapture Strength • Yield Strength • Tensile Strength • B

konstantin123 [22]

Answer: Rupture strength

Explanation: Rupture strength is the strength of a material that is bearable till the point before the breakage by the tensile strength applied on it. This term is mentioned when there is a sort of deformation in the material due to tension.So, rupture will occur before whenever there are chances of failing and the material is still able to bear stresses before failing.

7 0

2 years ago

A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given

viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar.

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0

3 years ago

When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b

laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

$\tau_{ave} = \frac{T}{2tA_{m}}$

$A_m$ = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, $A_m$ = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

$\tau_{ave} = \frac{T}{2tA_{m}}$

Plugging in then values, gives;

$\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa$

The shear stress will be 80,000,000 Pa or 80 MPa.

7 0

2 years ago