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3 years ago

Find all the words, Figure out my puzzle!

Engineering

2 answers:

Mandarinka [93]3 years ago

5 0

Answer:

I found the word!

It was Free points.

Alinara [238K]3 years ago

5 0

Did fake salt Felix turtle ninja Alice friend

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Mong m.n giúp mình vs ạ

Ivenika [448]

Answer:

see vous se to pe a he ko off a nack u

4 0

2 years ago

A liquid propellant engine has the following characteristics: chamber pressure of 7 MPa, constant ratio of specific heats of 1.3

Cloud [144]

Answer:

1.55
260 N.s
3370 m
1.6
43.75 kg/s

Explanation:

1) Thrust coefficient at sea level.

Cfsl = TSL / Pca

TSL = Mp * Vc + ( Pc - Pa )Ac

Mp = mass flux = 43.75 kg/s

∴ Cfsl = Mp Vc / Pca + ( Pc - Pa )/Pc * ( Ac / A* )

= 1.6 - 0.04923 = 1.55

<u>2) Specific impulse at sea </u>

Isp = Vc / g = 2549.75 / 9.81

= 260 N.s

3) Altitude at optimal expansion

H = 3370 m

<u>4) thrust coefficient at optimal expansion </u>

CF = 1.6

attached below is the detailed solution

<u>5) Mass flux through the throat </u>

Mass flux = P1 * At / Cc

= ( 7*10^6 * 0.01 ) / 1600

= 43.75 kg/s

8 0

2 years ago

Suppose that the weights for newborn kittens are normally distributed with a mean of 125 grams and a standard deviation of 15 gr

kherson [118]

(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.

<h3>Weight distribution of the kitten</h3>

In a normal distribution curve;

2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%

1 standard deviation (d) below the mean (M), (M - d) is at 16 %
1 standard deviation (d) above the mean (M), (M + d) is at 84%
2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%

M - 2d = 125 g - 2(15g) = 95 g

M - d = 125 g - 15 g = 110 g

95 g is at 2% and 110 g is at 16%

(16% - 2%) = 14%

(110 - 95) = 15 g

14% / 15g = 0.93%/g

From 95 g to 99 g:

99 g - 95 g = 4 g

4g x 0.93%/g = 3.72%

99 g will be at:

(2% + 3.72%) = 5.72%

Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

<h3>Weight of the kitten in the 90th percentile</h3>

M + d = 125 + 15 = 140 g (at 84%)

M + 2d = 125 + 2(15) = 155 g ( at 98%)

155 g - 140 g = 15 g

14% / 15g = 0.93%/g

84% + x(0.93%/g) = 90%

84 + 0.93x = 90

0.93x = 6

x = 6.45 g

weight of a kitten in 90th percentile = 140 g + 6.45 g = 146.45 g

Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g

Learn more about standard deviation here: brainly.com/question/475676

#SPJ1

7 0

1 year ago

Twenty six million gallons per day of wastewater with a DO of 1.00 mg/L is discharged into a river with a DO of 6.00 mg/L. If th

laiz [17]

Answer:

oxygen deficit = 3.851 mg/L

Explanation:

given data

flow rate of the river= 165 × $10^{6}$ gal/d

saturation value of dissolved oxygen = 9.17 mg/L

to find out

oxygen deficit he two flows

solution

we will apply here formula for dissolved oxygen content after dilution is

Do mix = $\frac{Qw*(Do)w +Qr*(Do)r}{Qw+Qr}$ ..........................1

here Qw is rate of flow of waste water i.e 26 × $10^{6}$ gal/d

(Do)w is Do of waste waster i.e 1 mg/L

Qr is aret of flow of river i.e 165 × $10^{6}$ gal/d

(Do)r is do of river water i.e 6 mg/L

so put all value in equation 1 we get

Do mix = $\frac{26*10^6*1 +165*10^6*6}{(26+165)10^6}$

solve we get

Do mix = 5.319 mg/L

oxygen deficit = saturation oxygen - (Do) mix ..............2

oxygen deficit = 9.17 - 5.319

oxygen deficit = 3.851 mg/L

8 0

2 years ago

Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at

AysviL [449]

Answer:

(a) The reversible work is 207 kJ/kg

(b) The irreversibility rate is -38.39 kJ/kg

Explanation:

State1 : p1 = 100kpa, T1= 25+273 =298k

From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK

State 2a:p2=2mpa,t2=540k (actual condition 2a)

h2a= 544.35 kJ/kg,s2a0=2.29906

actual work input to the compressor =wout=h1-h2+Qin

=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)

=(-246.17)kJ/kg(-ve sign indicates the work is given into the system

a) Reversible work= Win actual - any irreversiblities present

=246.17 + irreversibilty

b) irreversibility = T0(Entopy generation Sgen) for air, Sgen

=s20-s10-Rln(p2/p1), T0=250C

=(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)

= 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]

= -38.39 kJ/kg

a)Reversible work = Win actual -any irreversiblities present

=246.17 + irreversibilty

=246.17+-38.39

=207 kJ/kg

8 0

3 years ago