Using a definition of the limit of function, prove that the following functions are continuous

Engineering

1 answer:

Naddika [18.5K]2 years ago

3 0

Answer:

... you gotta do divson?

Explanation:

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___________ is NOT a common injury that an automotive tech may experience at work.

Degger [83]

Answer:The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

Explanation:

The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

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3 years ago

A commercial facility has a three-phase, 277 volt, 1200 Amp service. What is the required nominal transformer size needed for th

alekssr [168]

Answer:

The answer is "Option e".

Explanation:

Given value:

$\to v_{ph}=277 \ \ \ \ \ \ \ I_{ph} =1200\\\\$

$vA= v_{Ph} I_{ph} \ \ \ \to for \ single \ phase\\\\=3v_{ph} I_{ph} \ \ \ \ \ \to for \ 3 - \phi \\\\s = 3\times 277 \times 1200 \\\\$

$=997.200 \ K VA\\\\= 1000 \ k VA$

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3 years ago

A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto

vazorg [7]

Answer:

$(M_t)_{rated}=61.11lb-in$

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = $4 \times 0.7457$ =2.9828 KW

service factor(k)= 2.75

now,

$KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }$