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Nutka1998 [239]
3 years ago
11

True or False:A retractable service pit cover can help other shop employees from falling into the pit while another technician i

s in the pit servicing a car
Engineering
2 answers:
coldgirl [10]3 years ago
7 0

Answer:

f

Explanation:

elena-s [515]3 years ago
6 0

A retractable service pit cover can help other shop employees from falling into the pit while another technician is in the pit servicing a car? The given statement is true statement.

<u>Explanation:</u>

This cover prevent the employees from the accidental fall into the inspection pit and also provides access to underside of the vehicle. It is Easy to operate and has fast response to employ and deploy. It keeps the staff and visitors too safe.

The technician can open it to any position, which provide either full or partial exposure to the vehicle for servicing. It can handle to up to 200kg, where the technicians can stand on it to service the front or rear part of the vehicle.

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An asphalt concrete mixture includes 94% aggregates by weight. The specific gravity of aggregate and asphalt are 2.7 and 1.0, re
riadik2000 [5.3K]

Answer:

The correct solution is "5.74%".

Explanation:

The given values are:

Gravity of aggregate,

G_{agg}=2.7

Gravity of asphalt,

G_{asp}=1.0

Asphalt concrete mixture,

W_{agg}=0.94 \ W_m

We know that,

W_{asp}=W_m-W_{agg}

        =0.06 \ W_m

Now,

The theoretical specific gravity of mix,

⇒ G_t=\frac{W_{agg}+W_{asp}}{\frac{W_{agg}}{G_{agg}} +\frac{W_{asp}}{G_{asp}} }

By putting the values, we get

         =\frac{0.94 \ Wm+0.06 \ Wm}{\frac{0.94 \ Wm}{2.7} +\frac{0.06 \ Wm}{1} }

         =2.45

hence,

The percentage of voids will be:

⇒  %V = \frac{G_t-G_m}{G_t}\times 100

           = \frac{2.45-2.317}{2.45}\times 100

           = \frac{0.133}{2.317}\times 100

           = 5.74 (%)  

8 0
3 years ago
1 UNREAD MESSAGE
myrzilka [38]

Answer:

triangular trade

Explanation:

triangular trade

7 0
2 years ago
Con que otro nombre se le conoce a los delitos informaticos
kow [346]
Repuesto: Cyberdelito
7 0
3 years ago
You are to assess the biomechanics of a male’s arm using his bicep to hold a 20 kg object in his hand. The upper arm is perpendi
cestrela7 [59]

Answer:

Explanation:

The detailed analysis, with free body diagram and step by step calculations with appropriate substitution is as shown in the attached files.

5 0
4 years ago
Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of th
sleet_krkn [62]

Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, \sigma is given by

\sigma=\frac {F}{A} where F is force and A is area and area is given by \frac {\pi d^{2}}{4} hence

\sigma=\frac {4F}{\pi d^{2}} where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

\sigma=\frac {4*9970 N}{\pi 0.01m^{2}}=126941982.6 N/m^{2}

\sigma \approx 127 Mpa

From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

\epsilon=\frac {\triangle l}{l} where\epsilon is the strain, \triangle l is elongation and l is original length and making elongation the subject

\triangle l= \epsilon \times l and substituting strain with 0.0015 and length l with 100.6 mm then

\triangle l=0.0015\times 100.6=0.1509 mm

(b)

Lateral strain is given by

\epsilon_{lat}=\frac {\triangle d}{d} and substituting -v\epsilon for \epsilon_{lat} where v is poisson ratio then

-v\epsilon=\frac {\triangle d}{d} and making \triangle d the subject then

\triangle d=-vd\epsilon and substituting 0.35 for v, 0.0015 for strain and 10 mm for d

\triangle d=-(0.35)*10*0.0015=-0.00525 mm and the negative sign indicates decrease in diameter

8 0
3 years ago
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