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3 years ago

Which of the following are considered software piracy? Check all of the boxes that apply.

Engineering

1 answer:

Serga [27]3 years ago

7 0

Answer:

The answer is copying a program to give to someone else to use

and burning a copy of a DVD to sell

Explanation:

Software piracy is the act of stealing software that is legally protected. This stealing includes copying, distributing, modifying or selling

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What are the best collages for architectural learning?

Sati [7]

Answer:

I don't where you live, but I named few around the world. Some are Virgi..nia Tech, Unive..rsity of Flor..dia, and Oklah..oma Stat..e Univ..ersity.

4 0

2 years ago

The water behind Hoover Dam is 206m higher than the Colorado river below it. At what rate must water pass through the hydraulic

Leokris [45]

Answer:

m' = 4948.38 kg/s

Explanation:

For a case of 100% efficiency, the power produced must be equal to the rate of potential energy conversion

GIVEN THAT

Power = 100 MW

rate of Potential energy = (m')*g*h

100*10^6 = (m')*9.81*206

m' = 4948.38 kg/s

3 0

3 years ago

Read 2 more answers

Problem 3: Soil Classification using the AASHTO and USCS Systems

nataly862011 [7]

<u>Solution:</u>

$Given\\ $\quad W=3000 Ib , \quad m=\frac{W}{g}=\frac{3000}{322} \ slug =93.1677 slug$\\K_{e q}=2160 lbs / wp =2100 \frac{ lbs }{10} \frac{ x 12}{1 ft }=(2160 \times 12) lb / ft$$$

a) The natural frequency

$\begin{aligned}&\left(\omega_{n}\right)=\sqrt{\frac{K_{e q}}{m}}\\&=\sqrt{\frac{2160 \times 12}{93.1677}}\\&\omega_{n}=16.68 \text { rad } | s\\&\omega_{n}=\frac{2 \pi}{T}\\&16.68=\frac{2 \pi}{T}\\&T=0.3767 s\end{aligned}$

$Given, $t=10 s , \quad y(t)=6 in = A$\\$y(t)=A \cos \left(\omega_{n} t+\phi\right) \rightarrow 0$\\$6=6 \cos (16.68 \times 10+\phi)$\\$1=\cos (166.8+\phi)$\\$166.8+\phi=0$\\\phi=-166.8\)\\At $t=0, \quad y(0)=6 \cos (16.68 \times 0-166.8)$ {y(0)}=-5.74 in$

5 0

2 years ago

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$