Print reading for industry unit 9 review questions

Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have fou

Bingel [31]

Ucsaaaaauxx627384772938282’cc ed un e uff ridicolizzarla +golfista

4 0

3 years ago

The Reynolds number is a dimensionless group defined for a fluid flowing in a pipe as Re Durho/μ whereD is pipe diameter, u is f

Gala2k [10]

Answer:

the flow is turbulent

Explanation:

The Reynolds number is given by

Re=ρVD/μ

where

V=fluid speed=0.48ft/s=0.146m/s

D=diameter=2.067in=0.0525m

ρ=density=0.805g/cm^3=805Kg/m^3

μ=0.43Cp=4.3x10^-4Pas

Re=(805)(0.146)(0.0525)/4.3x10^-4=14349.59

Re>2100 the flow is turbulent

Note: if you do not want to use a calculator you can use the graphs to calculate the Reynolds number according to their properties

8 0

2 years ago

Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equat

PtichkaEL [24]

Answer:

Detailed solution is attached below in three simple steps the problem is solved.

7 0

2 years ago

A man releases a stone (at rest, Vo=0) from the top of a tower. During the last second of its travel, the stone falls through a

RideAnS [48]

Answer:

Height of tower equals 122.5 meters.

Explanation:

Since the height of the tower is 'H' the total time of fall of stone 't' is calculated using second equation of kinematics as

Since the distance covered in last 1 second is $\frac{9H}{25}$ and the total distance covered in 't' seconds is 'H' thus the distance covered in the first (t-1) seconds of the motion equals

$S_{t-1}=S_{t}-S_{last}\\\\S_{t-1}=H-\frac{9H}{25}=\frac{16H}{25}$

Now by second equation of kinematics we have

$S=ut+\frac{1}{2}gt^{2}\\\\S=\frac{1}{2}gt^{2}(\because u=0)$

Thus we have

$\frac{16H}{25}=\frac{1}{2}g(t-1)^{2}.............(i)\\\\H=\frac{1}{2}gt^{2}..............(ii)$

Dividing i by ii we get

$\frac{16}{25}=\frac{(t-1)^{2}}{t^2}\\\\\therefore \frac{t-1}{t}=\frac{4}{5}\\\\\therefore t=5secs$

Thus from equation ii we obtain 'H' as

$H=\frac{1}{2}\times 9.8\times 5^{2}=122.5meters$

6 0

2 years ago

I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have

drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = $\lambda e^{-\lambda t}$

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

$P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1$

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

$p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905$

(c) The Poisson distribution is presented as follows;

$P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!} = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!} = 3.915\times 10^{-12}$

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) $CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}$ .

4 0

3 years ago