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s2008m [1.1K]
3 years ago
13

6. Question

Engineering
1 answer:
valkas [14]3 years ago
4 0

Answer:

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

Explanation:

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Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i <= N; i = (i*2)+17 ) for(k = i
MissTica

Answer:

Alice is correct.

The loop are dependent.

Explanation:

for(i = 1; i <= N; i = (i*2)+17 )

for(k = i+1; k <= i+N; k = k+1) // notice i in i+1 and i+N

printf("B")

This is a nested for-loop.

After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.

So, the second for-loop depend on the first for loop to get the value of i. For clarity purpose, code indentation or use of curly brace is advised.

8 0
3 years ago
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Hello it's my new id<br>I am numu ​
Sonja [21]

Answer:

i am felix

Exp  lanation:

nice to meet you

6 0
3 years ago
The “Sun-Star” Company has purchased new office furniture for their offices at a retail price of $100,000. An additional $12,000 h
Zinaida [17]

Based on the cost of the furniture, the following are true:

  • a. $10,400.
  • b. $93,600
  • c. $20,800

<h3>Depreciation in second year</h3>

Depreciation per year = (Cost of furniture - Salvage value) / Useful life

Cost will include both the purchase price and the charge for insurance and shipping.

= (100,000 + 12,000 - 8,000) / 10

= $10,400

<h3>BV at end of first year</h3>

= Cost - Depreciation

= 104,000 - 10,400

= $93,600

<h3>BV after 8 years </h3>

= Cost - (Depreciation x 8 years )

= 104,000 - (10,400 x 8)

= $20,800

In conclusion, depreciation is $10,400 per year.

Find out more about SL depreciation at brainly.com/question/13734742.

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
3 years ago
Oil is graded by the
Angelina_Jolie [31]

Answer:

Oil is graded by the Society of Automotive Engineers - B.

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3 years ago
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