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vivado [14]
3 years ago
14

2.14 (a) Using series/parallel resistance reductions, find the equivalent resistance between terminals A and B in the circuit of

Figure P2.14. (b) Repeat, but with C and D shorted together.
Engineering
1 answer:
olasank [31]3 years ago
5 0

Answer:

There is no attachment

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Table 1(a) shows the marks obtained by 40 students in an examination
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I dont know the answer to this
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3 years ago
Another focus of effective communication, according to Stephen Covey, is ensuring that:
svetoff [14.1K]

Answer:

Stephen Covey believes this principle is the key to effective interpersonal communication. Seek first to understand, then to be understood. This habit is about communicating with others. It's about developing the habit of listening carefully and really understanding the other person BEFORE giving your thoughts.

Explanation:

4 0
3 years ago
Read 2 more answers
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
What technology has been used for building super structures​
AleksAgata [21]

Answer: Advanced technologixal machines

Explanation: such as big cranes, multiple workers helping creat said structure, and big bull dozers

7 0
3 years ago
A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat
ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000  

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n  ..........1

and

total cost of capital intensive  = $1,244,000 + $92.5 n   ..........2

so here in both we prefer cost of capital if cost of capital intensive less than cost of labor

$1,244,000 + $92.5 n  <  $33800 + $143 n

solve we get

n > \frac{906000}{50.5}

n > 17941

and

cost of producing less than selling cost so here

$1,244,000 + $92.5 n < 197 n

solve it we get

n > \frac{1244000}{104.5}  

n > 11904

so in both we get greatest no is 17941

so no of unit is 17941

3 0
3 years ago
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