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2 years ago

Which of the following is likely to have the suffix "" after the domain name in its URL?

Engineering

2 answers:

vovangra [49]2 years ago

5 0

Answer:

Microsoft is the correct answer

Leviafan [203]2 years ago

4 0

Question:

Which of the following is likely to have the suffix ".com" after the domain name in its URL?

The World Wide Fund for Nature Massachusetts

Institute of Technology

Microsoft

The Federal Reserve

UNICEF

Answer:

Microsoft

Explanation:

URL (universal resource locator): is the address of an internet web page. The address also contains the domain name of the web page, a domain name is used to identify the address of a particular website on the internet, a domain name can be generic (.com, .net, .org, .info), sponsored (.edu, .gov, .int), or country code(.us, .uk, .au) domain. Therefore an individual, company, organization or country is addressed the type of domain name entity they belong to.

Microsoft is likely to have a “.com” domain main suffix because it is a profit making company. Others are government and non-profit making organization which are likely to either have .gov and .org as domain names.

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Compute L, T, M, LC, and R and stations of the BC and EC for the circular curve with the given data of: I (delta) = 22°15′00" an

Mars2501 [29]

Answer:

L = 475.718

T = 240.89 ft

M = 23.0195

LC = 472.728

R = 1225 ft

Explanation:

See the attached file for the calculation.

8 0

2 years ago

What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00

yKpoI14uk [10]

Answer:

The velocity of flow is 10.0 m/s.

Explanation:

We shall use Manning's equation to calculate the velocity of flow

Velocity of flow by manning's equation is given by

$V=\frac{1}{n}R^{2/3}S^{1/2}$

where

n = manning's roughness coefficient

R = hydraulic radius

S = bed slope of the channel

We know that for an asphalt channel value of manning's roughness coefficient = 0.016

Applying values in the above equation we obtain velocity of flow as

$V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s$

7 0

2 years ago

Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba

Tpy6a [65]

Answer:

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Explanation:

The optimum cutting speed for the minimum cost

$V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)$

Where,

C,n = Taylor equation parameters

$T_h$ =Tool changing time in minutes

$C_e$ =Cost per grinding per edge

$C_m$ = Machine and operator cost per minute

On comparing with the Taylor equation $VT^n=C$ ,

Tool life,

$T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)$

Given that,

Cost of operator and machine time $=\$40/hr=\$0.667/min$

Batch setting time = 2 hr

Part handling time: $T_h=2.5$ min

Part diameter: $D=73 mm$ $=73\times 10^{-3} m$

Part length: $l=250 mm=250\times 10^{-3} m$

Feed: $f=0.30 mm/rev= 0.3\times 10^{-3} m/rev$

Depth of cut: $d=3.5 mm$

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, $C_e=$ $\$20/15+2=\$3.33/edge$

Tool changing time, $T_t=3$ min.

$C= 80 m/min$

$n=0.130$

(a) From equation (i), cutting speed for the minimum cost:

$V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}$

$\Rightarrow 47.7$ m/min

(b) From equation (ii), the tool life,

$T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}$

$\Rightarrow T=53.4$ min

(c) Cycle time: $T_c=T_h+T_m+\frac{T_t}{n_p}$

where,

$T_m=$ Machining time for one part

$n_p=$ Number of pieces cut in one tool life

$T_m= \frac{l}{fN}$ min, where $N=\frac{V_{opt}}{\pi D}$ is the rpm of the spindle.

$\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc$

So, the number of parts produced in one tool life

$n_p=\frac {T}{T_m}$

$\Rightarrow n_p=\frac {53.4}{4.01}=13.3$

Round it to the lower integer

$\Rightarrow n_p=13$

So, the cycle time

$T_c=2.5+4.01+\frac{3}{13}=6.74$ min/pc

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc$

(e) Total time to complete the batch= Sum of setup time and production time for one batch

$=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times4.01}{457}=0.4387=43.87\%$

Now, for the cemented carbide tool:

Cost per edge,

$C_e=$ $\$8/6=\$1.33/edge$

Tool changing time, $T_t=1min$

$C= 650 m/min$

$n=0.30$

(a) Cutting speed for the minimum cost:

$V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min$ [from(i)]

(b) Tool life,

$T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min$ [from(ii)]

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc$

$n_p=\frac {7}{0.53}=13.2$

$\Rightarrow n_p=13$ [ nearest lower integer]

So, the cycle time

$T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc$

(e) Total time to complete the batch $=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.53}{275.5}=0.0962=9.62\%$

Similarly, for the ceramic tool:

$C_e=$ $\$10/6=\$1.67/edge$

$T_t-1min$

$C= 3500 m/min$

$n=0.6$

(a) Cutting speed:

$V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}$

$\Rightarrow V_{opt}=2105 m/min$

(b) Tool life,

$T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min$

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc$

$n_p=\frac {2.33}{0.091}=25.6$

$\Rightarrow n_p=25 pc/tool\; life$

So,

$T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc$

(e) Total time to complete the batch

$=2\times60+ {50\times 2.63}=251.5 min=4.19 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.091}{251.5}=0.0181=1.81\%$

3 0

3 years ago

How should you move your board through the planer? (Pick two choices.)

iragen [17]

Answer:

I would say do it at an even pace

Explanation:

Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast

4 0

2 years ago

What should you, the worker, be aware of with regard to evacuation procedures at your workplace

Alinara [238K]

Answer:

As a worker, it is important to follow the proper set of instructions or emergency plans during an emergent situation. Not carefully following the rules may result to a bigger problem such as further injury and damage to property.

Explanation:

Evacuation Procedure- This is a step-by-step procedure that people follow in order to safely vacate any building or place. This procedure is applicable to any situation, such as the workplace. This is now called the <em>Workplace Evacuation Procedure. </em>This is very important because there are so many unpredictable situations or events that are happening in the world right now, such as fire or earthquake. This procedure is being done through an evacuation plan.

The awareness of the workers regarding the proper way to evacuate during emergency situation is very important. It will be easier for them to know where to locate the nearest exit route. They will also learn to stop any form of device or equipment that could cause a hazzard during the situation. In case of the hospital, which is also a workplace, the employees will also learn how to assist the patients before themselves. They will also know where to assemble if there's a need to do so.

7 0

2 years ago

Read 2 more answers