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Black_prince [1.1K]
3 years ago
14

Words and numbers can be printed using many different or type styles.

Engineering
1 answer:
Nitella [24]3 years ago
8 0
I believe the answer is


Appendix

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before adjusting drive-belt tension, technician a checks for proper pulley alignment. technician b looks up the specified belt t
Vsevolod [243]

Answer:

Technician b is correct

Explanation:

Before adjusting drive-belt tension, it is very important to check the vehicle workshop manual for specified belt tension, so that you can match your reading against the specification in the vehicle's service manual. If the tension reading you have matches the suggested reading in the vehicle's service manual and the belt is not damaged then you do not need to proceed any further. But if the reading does not match, then you can adjust the belt tension.

Therefore, technician b is correct.

5 0
3 years ago
Who plays a role in the financial activities of a company?
KatRina [158]

Hey,

Who plays a role in the financial activities of a company?

<em>O D. Everyone at the company, including managers and employees</em>

3 0
4 years ago
A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream
Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}

Substituting figures in the equation we have:

= \frac{1}{1+0.1(2.5-1)+0(2-1)}

= 0.75

Let's now calculate equivalent flow rate of the car using:

Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}

= \frac{7500}{0.9*4*0.75*1}

= 2777.7 pc/h/en

Calculating for traffic density, we have:

D = \frac{Vp}{FFS}

D = \frac{2777.7}{70}

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0
4 years ago
Which option identifies the tool best to use in the following scenario?
Lena [83]

Answer:

an Allen wrench

Explanation:

it is hexagonal

3 0
3 years ago
What are some things that AREN’T manufactured
snow_lady [41]
Ray Bans
Arrow Shirts
Gillette
3 0
3 years ago
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