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VARVARA [1.3K]
3 years ago
8

You are traveling upstream on a river at dusk. You see a buoy with the number 5 and a flashing green light . What should you do?

Engineering
1 answer:
nalin [4]3 years ago
8 0

Answer:

please give brainliest my brother just got the corona virus

Explanation:

this is my brothers account he wants to get 5 brainliest

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A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory
Molodets [167]

Answer:

Explanation:

Mean temperature is given by

T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}

Tmean = (Ti + T∞)/2

T_mean = 107.5^{0}

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

\mu = 221.6 × 10⁻⁷N.s/m²

\kappa = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α  = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

              = 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/\mu

Substituting for Pv

Re = 4m/(πD\mu)

     = (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

     = 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0
3 years ago
Which two is right about febuary 14
igor_vitrenko [27]

Answer:A and B

Explanation:

3 0
3 years ago
Read 2 more answers
PLS HELP ME
Oksana_A [137]

Answer:

The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.

Explanation:

160 - 120 = 40

120 = 100

40 = X

40 x 100 / 120 = X

4000 / 120 = X

33.333 = X

120 = 100

160 = X

160 x 100 /120 = X

16000 / 120 = X

133.333 = X

4 0
3 years ago
A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at
Radda [10]

Answer:

  • <u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.

4 0
3 years ago
What must engineers keep in mind so that their solutions will be appropriate?
vekshin1

Answer:

Context

Explanation:

It is of great value for an engineer to keep the context of his/her experiment in mind.

7 0
3 years ago
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