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Korolek [52]
3 years ago
6

Let S = { p q |p, q are prime numbers greater than 0} and E = {0, −2, 2, −4, 4, −6, 6, · · · } be the set of even integers. . Pr

ove that |S| = |E| by constructing a bijection from S to E.
Engineering
1 answer:
Finger [1]3 years ago
7 0

Answer:

prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S

Explanation:

Given that S = { p q |p, q are prime numbers greater than 0}

                    E = {0, −2, 2, −4, 4, −6, 6, · · · }

To prove  by constructing a bijection from S to E

detailed  solution attached below

After the bijection :

<em>prove that | S | = | E |</em> :  every element of S there is an Image on E , while not every element on E has an image on S

∴ we can say sets E and S are infinite sets

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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636
luda_lava [24]

Answer:

The original length of the specimen is found to be 76.093 mm.

Explanation:

From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:

Initial Volume = Final Volume

πd1²L1/4 = πd2²L2/4

d1²L1 = d2²L2

L1 = d2²L2/d1²

where,

d1 = initial diameter = 19.636 mm

d2 = final diameter = 19.661 mm

L1 = Initial Length = Original Length = ?

L2 = Final Length = 75.9 mm

Therefore, using values:

L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²

<u>L1 = 76.093 mm</u>

5 0
3 years ago
An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter
slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg ,  1980.39 MJ/Kmol

Explanation:

molar mass of  ethanol (C2H5OH) = 46 g/mol

molar mass of   octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol =  0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV  of  ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV  of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol)  for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value  ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV  in MJ/kg  = 0.85 * 29.7 + 0.15 * 47.9  = 32.43 MJ/kg

HHV in  MJ /kmol  =  0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0
3 years ago
Example for strengthening mechanism in single-phase material A. Dispersion strengthening B. Precipitation hardening C. Fiber str
Lena [83]

Answer:

Example for strengthening mechanism in single-phase material: Strain hardening- D.

3 0
3 years ago
Read 2 more answers
The conditions at the beginning of compression in an Otto engine operating on hot-air standard with k=1.35 and 101.325 kPa, 0.05
Katyanochek1 [597]

Answer:

P_m=181.42 KPa

Explanation:

P_1=101.325 KPa,V_1=0.05m^3,T_1=32C,K=1.35

Clearance is 8%.

Heat added=15 KJ

We know that compression ratio r=1+\dfrac{1}{C}

r=1+\dfrac{1}{0.08}  

r=13.5

r=\dfrac{V_1}{V_2}

13.5=\dfrac{0.05}{V_2}

V_2=3.7\times 10^{-3}

We know that efficiency of otto cycle

\eta =1-\dfrac{1}{r^{k-1}}

\eta =1-\dfrac{1}{13.5^{1.35-1}}

\eta =0.56

\eta =\dfrac{W}{Q}

W is the work out put and Q is the heat addition.

0.56 =\dfrac{W}{15}

W=8.4 KJ

We know that Work =Mean effective pressure x swept volume.

Here swept volume V_s=V_1-V_2

V_s=0.05-3.7\times 10^{-3}

V_s=0.0463 m^3

Noe by putting the values

Work =Mean effective pressure x swept volume.

8.4=P_m\times 0.0463

P_m=181.42 KPa

6 0
3 years ago
Why is tubing sometimes coiled when installed in a car or vechile
andriy [413]

Answer:

Coiled tubing is often used to carry out operations similar to wire lining.

8 0
3 years ago
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