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3 years ago

Can someone please do this for me ive been sick for the past week and am behind

Engineering

1 answer:

AlekseyPX3 years ago

7 0

Answer:

We didn’t choose one

Explanation:A recent survey of directors, CEOs, and senior executives found that digital transformation (DT) risk is their #1 concern in 2019. Yet 70% of all DT initiatives do not reach their goals. Of the $1.3 trillion that was spent on DT last year, it was estimated that $900 billion went to waste. Why do some DT efforts succeed and others fail?

Fundamentally, it’s because most digital technologies provide possibilities for efficiency gains and customer intimacy. But if people lack the right mindset to change and the current organizational practices are flawed, DT will simply magnify those flaws. Five key lessons have helped us lead our organizations through digital transformations that succeeded.

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Use pseudocode. 1) Prompt for and input a saleswoman's sales for the month (in dollars) and her commission rate (percentage). Ou

aleksandr82 [10.1K]

Answer:

The pseudocode is as follows:

Input SalesAmount

Input CommissionRate

CommissionEarned= SalesAmount * (CommissionRate/100)

Print CommissionEarned

Explanation:

This gets input for SalesAmount

Input SalesAmount

This gets input for CommissionRate

Input CommissionRate

This calculates the CommissionEarned

CommissionEarned= SalesAmount * (CommissionRate/100)

This prints the calculated CommissionEarned

Print CommissionEarned

3 0

3 years ago

Why does teachers grade things that are not due yet

pav-90 [236]

I think because if you’ve already turned it in they might as well grade asap instead of waiting

4 0

3 years ago

Read 2 more answers

A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para

mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c) 54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

$V_p=\frac{3810.5}{\sqrt{3} }=2200 V$

The complex power S is given as:

$S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA$

$where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA$

The line current I is given as:

$I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A$

The phase voltage at the sending end is:

$V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV$

The magnitude of the line voltage at the source end of the line ( $V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V$

b) The Total real and reactive power loss in the line is:

$S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000$

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

$S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000$

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0

4 years ago

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio

Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

$v=23.6m/s$