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Gnoma [55]
2 years ago
9

Which of the following is a core component of the Ford motor company greener and smarter strategy

Engineering
1 answer:
Andreas93 [3]2 years ago
8 0

Answer:

Ford's comprehensive global sustainability strategy includes a focus on the development of environmentally friendly vehicle technologies, including hybrids, diesels, bio-diesels, advanced engines and transmissions, E85 ethanol, plug-in hybrids, hydrogen internal combustion engines and fuel cells.

Explanation:

iv'e read about it

You might be interested in
A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t
Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa

The design factor for metals is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}

The design factor for plastic and ceramics is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5

This material will break.

3 0
2 years ago
Problem 5) Water is pumped through a 60 m long, 0.3 m diameter pipe from a lower reservoir to a higher reservoir whose surface i
kap26 [50]

Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2

where P_1 = P_2 = 0

V1 AND V2  =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV  =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f  = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

4 0
2 years ago
Why is it so dangerous to use a ground lift on a metal cased power tool
MaRussiya [10]
Answer:

Removal of the safety ground connection on equipment can expose users to an increased danger of electric shock and may contradict wiring regulations. ... If a fault develops in any line-operated equipment, cable shields and equipment enclosures may become energized, creating an electric shock hazard.
7 0
1 year ago
A railcar with an overall mass of 78,000 kg traveling with a speed vi is approaching a barrier equipped with a bumper consisting
sergij07 [2.7K]

Answer:

v₀ = 2,562 m / s  = 9.2 km/h

Explanation:

To solve this problem let's use Newton's second law

              F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v

              F dx = m v dv

We replace and integrate

            -β ∫ x³ dx = m ∫ v dv

            β x⁴/ 4 = m v² / 2

We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max

        -β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)

         x_max⁴ = 2 m /β   v₀²

         

Let's look for the speed that the train can have for maximum compression

         x_max = 20 cm = 0.20 m

         

         v₀ =√(β/2m)   x_max²

Let's calculate

          v₀ = √(640 106/2 7.8 104)    0.20²

          v₀ = 64.05  0.04

          v₀ = 2,562 m / s

          v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)

          v₀ = 9.2 km / h

5 0
2 years ago
A 60-kg woman holds a 9-kg package as she stands within an elevator which briefly accelerates upward at a rate of g/4. Determine
Anuta_ua [19.1K]

Answer:

force R = 846.11 N

lifting force L = 110.36 N

if cable fail complete both R and L will be zero

Explanation:

given data

mass woman mw = 60 kg

mass package mp = 9 kg

accelerates rate a = g/4

to find out

force R and lifting force L and if cable fail than what values would R and L acquire

solution

we calculate here first reaction R force

we know elevator which accelerates upward

so now by direction of motion , balance the force that is express as

R - ( mw + mp ) × g = ( mw + mp ) × a

here put all these value and a = g/4 and use g = 9.81 m/s²

R - ( 60 + 9 ) × 9.81 = ( 60 + 9  ) × g/4

R = ( 69  ) × 9.81/4  + ( 69 ) 9.81

R = 69  ( 9.81 + 2.4525 )

force R = 846.11 N

and

lifting force is express as here

lifting force = mp ( g + a)

put here value

lifting force = 9 ( 9.81 + 9.81/4)

lifting force L = 110.36 N

and

we know if cable completely fail than body move free fall and experience no force

so both R and L will be zero

5 0
2 years ago
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