How does concurrent engineering speed up product development?

A. The product and process are designed using computer-aided design.
B. The product and process are designed in series.
C. The product and process are designed in separate spaces.
D. The product and process are designed simultaneously.

Answer:

Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point). road

Explanation:

Answer:

1. 1111.5MPa

2. 56.1GPa

Explanation:

1. Longitudinal tensile stress can be obtained by obtaining the strength and volume of the fiber reinforcement. The derived formula is given by;

σcl = σm (1 - Vf) + σfVf

Substituting the figures, we will have;

45(1 - 0.30) + 3600(0.30)

45(0.70) + 1080

31.5 + 1080

= 1111.5MPa

2. Longitudinal modulus of elasticity or Young's modulus is the ability of an object to resist deformation. The derived formula is given by;

Ecl = EmVm + EfVf

Substituting the formula gives;

= 2.4 (1 - 0.30) + 131 (0.30)

= 2.4(0.70) + 39.3

= 16.8 + 39.3

= 56.1GPa

Using the appropriate relation, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.

Longitudinal tensile stress can be obtained using the relation :

Substituting the values into the relation:

45(1 - 0.30) + 3600(0.30)

45 × 0.70 + 1080

31.5 + 1080

= 1111.50 MPa

2.)

Longitudinal modulus of elasticity is obtained using the relation :

Substituting the values thus :

2.4 (1 - 0.30) + 131 (0.30)

= 2.4 × 0.70 + 39.3

= 16.8 + 39.3

= 56.10 GPa

Hence, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.

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Answer:

a. [tex]L_o[/tex]  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, [tex]L_o[/tex]  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

[tex]L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)[/tex]

For the school, [tex]K_{LL}[/tex] = 2

Therefore, we have;

[tex]L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf[/tex]

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, [tex]W_d[/tex] = b × [tex]W_{D + L}[/tex] =

∴  [tex]W_d[/tex] =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, [tex]W_d[/tex] = 812.8 ft./lb

d. For the uniformly distributed load, we have;

[tex]V_{max}[/tex] = 812.8 × 26/2 = 10566.4 lbs

[tex]M_{max}[/tex] =  812.8 × 26²/8 = 68,681.6 ft-lbs

[tex]v_{max}[/tex] = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

[tex]P_L[/tex] = 1000 lb

[tex]W_{D}[/tex] = 20 × 16 = 320 lb/ft.

[tex]V_{max}[/tex] = 1,000 + 320 × 26/2 = 5,160 lbs

[tex]M_{max}[/tex] =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

[tex]v_{max}[/tex] = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

Answer:

Rocks

Explanation:

I am not sure tho bc they are made out of coal and I think coal is a kind of rock

Answer:

I'm taking a guess for ya I shall say Metal or and Minerals

Explanation:

Good luck