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4 years ago

13

3. What group is responsible for creating the product features that meet customer's needs and wants? A. Manufacturing B. Product

design C. Quality D. Engineering

1 answer:

AnnZ [28]4 years ago

6 0

Answer:

Explanation:

i think it's B

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Which of the following is a possible unit of ultimate tensile strength?

levacccp [35]

Answer:

Newton per square meter (N/m2)

Explanation:

Required

Unit of ultimate tensile strength

Ultimate tensile strength (U) is calculated using:

$U = \frac{Ultimate\ Force}{Area}$

The units of force is N (Newton) and the unit of Area is m^2

So, we have:

$U = \frac{N}{m^2}$

or

$U = N/m^2$

<em>Hence: (c) is correct</em>

4 0

3 years ago

valkas [14]

Answer:

Check the 2nd, 3rd and 4th statements.

Explanation:

4 0

3 years ago

A solid steel shaft has to transmit 100 kW at 160 RPM. Taking allowable shear stress at 70 Mpa, find the suitable diameter of th

MA_775_DIABLO [31]

Answer:

The diameter of the shaft is 80.5 mm.

Explanation:

Torsion equation is applied for the diameter of the solid shaft.

Step1

Given:

Power of the shaft is 100 kw.

Revolution per minute is 160 RPM.

Allowable shear stress is 70 Mpa.

Maximum torque is 20% more than the mean torque.

Step2

Mean torque is calculated as follows:

$P=T\omega$

$P=T(\frac{2\pi N}{60})$

$100\times 1000=T(\frac{2\pi 160}{60})$

T=5968.31 N-m

Step3

Maximum torque is calculated as follows:

$T_{max}=(1+\frac{20}{100})T$

$T_{max}=1.2T$

$T_{max}=1.2\times 5968.31$

T_{max}=7161.97 N-m

Step4

Apply torsional equation for diameter of shaft as follows:

$\tau _{max}=\frac{T_{max}}{\frac{\pi d^{3}}{16}}$

$70\times 10^{6}=\frac{7161.97}{\frac{\pi d^{3}}{16}}$

$d^{3}=5.211\times 10^{-4}$

d=0.0805 m

or,

d=80.5 mm

Thus, the diameter of the shaft is 80.5 mm.

7 0

4 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

4 years ago

Alicia had to get over her fear of heights in order to become comfortable maintaining the generators in wind turbines. professio

kow [346]

Answer:a

Explanation:

5 0

3 years ago