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3 years ago

Which bulb has the lowest total cost of operation? (a) Incandescent (b) Fluorescent (c) LED

Engineering

1 answer:

Finger [1]3 years ago

6 0

Answer: LED have the lowest cost of operation.

Explanation:

If we ignore the initial procurement cost of the items the operational cost of any device consuming electricity is given by

$Cost=Energy\times cost/unit$

Among the three item's LED consumes the lowest power to give the same level of brightness as compared to the other 2 item's thus LED's shall have the lowest operational cost.

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The viscosity of the water was 2.3×10^−5lb⋅⋅s/ft^2 and the water density was 1.94 slugs/ft^3. Estimate the drag on an 88-ft diam

Lina20 [59]

Answer:

hello your question is incomplete attached below is the complete question

answer : Drag force = 1.3 Ib

Explanation:

we have to represent the dimensions of the drag force in terms of FLT

i.e : D = f( d,v,p,u ) represented in terms of FLT

D = F , V = LT^-1, d = L, p = FL^-4 T^2

u = FL^-2 T, Number of independent terms = 5

attached below is the detailed solution

6 0

3 years ago

The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of the air

navik [9.2K]

Answer:

$Ma=\frac{260}{330} \\Ma=0.7878$

So, Ma < 1 Flow is Subsonic

Explanation:

Mach Number:

Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)

Mach < 1 Subsonic

Mach > 1 Supersonic

Ma= Speed of the object/Speed of the sound

$Ma=\frac{260}{330} \\Ma=0.7878$

So, Ma < 1 Flow is Subsonic

8 0

3 years ago

This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1

Ulleksa [173]

Answer:

L is not a regular language with formal proofs

Explanation:

(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails that L is a regular language. Then by the Pumping Lemma for Regular Languages, 

there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, 

s = xyz subject to the following conditions: 

(a) |y| > 0 

(b) |xy| ≤ p, and 

(c) ∀i > 0, xyi 

z ∈ L

(b) To determine that L is not a regular language, we mke use of proof by contradiction. lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,

The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject to the condtions as follows : 

(a) |y| > 0 

(b) |xy| ≤ p, and 

(c) ∀i > 0, xyi 

z ∈ L. 

Choose s = 0p10p 

. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.

for some k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus, xy0 

z should be in L. xy0 

z = xz = 0(p−k)10p 

It is shown that is is not in L. This is a contraption with the pumping lemma. our assumption that L is regular is incorrect, and L is not a regular language

6 0

3 years ago

⚠️Answer needed quick!!⚠️

Allisa [31]

Answer:

Explanation:

3 gifts

gift cost = x

$150 total to spend

$60 left after spending

5 0

3 years ago

Convert mechanical energy into electric energy. What can he use?

Nina [5.8K]

Answer:

<h2>Generator </h2>

Explanation:

A generator converts mechanical energy into electrical energy

7 0

3 years ago