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yKpoI14uk [10]
3 years ago
14

1. Two technicians are discussing tire rotation. Technician A says that you always follow the tire-rotation procedure outlined i

n the owner's manual or online service information. Technician B says that the modified "X" rotation pattern is seldom used. Which technician is correct? A. Both Technicians A and B B. Technician B only C. Neither Technician A nor B D. Technician A only ​
Engineering
1 answer:
siniylev [52]3 years ago
8 0

Answer:

don't know

Explanation:

huhuh

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The convection heat transfer coefficient for a clothed person standing in moving air is expressed as h 5 14.8V0.69 for 0.15 , V
Rom4ik [11]
Cychbjnivrxezyyihvhuytrruokjaa
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3 years ago
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*6–24. The beam is used to support a dead load of 400 lb>ft, a live load of 2 k>ft, and a concentrated live load of 8 k. D
lisabon 2012 [21]

Answer:

(a) maximum positive reaction at A = 64.0 k

(b) maximum positive shear at A = 32.0 k

(c) maximum negative moment at C = -540 k·ft

Explanation:

Given;

dead load  Gk = 400 lb/ft

live load Qk = 2 k/ft

concentrated live load Pk =8 k

(a) from the influence line for vertical reaction at A, the maximum positive reaction is

A_{ymax} = 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k

See attachment for the calculations of (b) & (c) including the influence line

3 0
3 years ago
From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displaye
Blababa [14]

Answer:

(a) Precipitation hardening

(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.

(2) The hardening/strengthening effect is not retained at elevated temperatures for this process.

(4) The strength is developed by a heat treatment.  

(b) Dispersion strengthening

(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.  

(3) The hardening/strengthening effect is retained at elevated temperatures for this process.

(5) The strength is developed without a heat treatment.  

7 0
3 years ago
A student lives in an apartment with a floor area of 60 m2 and ceiling height of 1.8 m. The apartment has a fresh (outdoor) air
USPshnik [31]

Answer:

4

Explanation:

5 0
2 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
3 years ago
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