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3 years ago

When a company in the United States employs people in India to answer their customer service calls this is an example of

Engineering

1 answer:

fgiga [73]3 years ago

3 0

Answer is : B Outsourcing.

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An escalator in a shopping center is designed to move 50 people, 75 kg each, at a constant speed of 0.6 m/s at 450 slope. Determ

REY [17]

Answer:

Power required to drive the escalator shall be equal to the rate at which the energies of the persons is increased.

$Energy=n\times mass\times g\times h\\\\\therefore Power=n\times mass\times g\times \frac{dh}{dt}\\\\Power=50\times 75\times 9.81\times 0.6sin(45)\\\\Power=15.607kW$

As we infer from the above equation If the velocity of the escalator is doubled then the Power required will also be doubled and become 31.215kW

3 0

3 years ago

A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto

likoan [24]

Answer:

2.455 W

Explanation:

The power dissipated in each branch is ...

P = V^2/R

So, the branch powers are ...

branch 1: 18^2/220 ≈ 1.473 W

branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

1.473 W + 0.982 W = 2.455 W

8 0

3 years ago

Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (1

WITCHER [35]

Answer:

Detailed solution is given below:

7 0

4 years ago

A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7

barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

$Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}$

$r=\dfrac{73.5}{232.45}$

r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

Now by putting the values

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

$tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }$ \

Ф=18.03°

So the shear angle is 18.03°.

4 0

4 years ago

Plz answer all of these questions!

statuscvo [17]

Answer: all you need to to is go to help me with this question.cm

Explanation:

5 0

3 years ago