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3 years ago

What is the missing number in the pattern below? 0, 1, 2, 3, 6, 11, 20, _?_, 68

Engineering

1 answer:

lbvjy [14]3 years ago

3 0

Answer:

0, 1, 2, 3, 6, 11, 20, 37, 68

Explanation:

Each number in the series, starting with 3, is the total of the three numbers before it. The following number is the total of the preceding three numbers, according to the pattern. The pattern's next number is 125.

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An automated transfer line is to be designed. Based on previous experience, the average downtime per occurrence = 5.0 min, and t

IRINA_888 [86]

Answer:

a) 28 stations

b) Rp = 21.43

E = 0.5

Explanation:

Given:

Average downtime per occurrence = 5.0 min

Probability that leads to downtime, d= 0.01

Total work time, Tc = 39.2 min

a) For the optimum number of stations on the line that will maximize production rate.

Maximizing Rp =minimizing Tp

Tp = Tc + Ftd

$= \frac{39.2}{n} + (n * 0.01 * 5.0)$

$= \frac{39.2}{n} + (n * 0.05)$

At minimum pt. = 0, we have:

dTp/dn = 0

$= \frac{-39.2}{n^2} + 0.05 = 0$

Solving for n²:

$n^2 = \frac{39.2}{0.05} = 784$

$n = \sqrt{784} = 28$

The optimum number of stations on the line that will maximize production rate is 28 stations.

b) $Tp = \frac{39.2}{28} + (28 * 0.01 * 5)$

Tp = 1.4 +1.4 = 2.8

The production rate, Rp =

$\frac{60min}{2.8} = 21.43$

The proportion uptime,

$E = \frac{1.4}{2.8} = 0.5$

3 0

3 years ago

For automobile engines, identify the best choice of motor oil.

Vlad1618 [11]

Answer:

C. Low viscosity for Maine in the winter and high viscosity for Florida in the summer.

Explanation:

The choice of motor oil or engine oil depends on the type of engine and also on the type of climate on which the engine runs.

In winter climates, we should use oil with low viscosity as at lower temperatures, the it flows more readily if its viscosity is low.

But the summer season needs a relatively thicker oil so as to reduce the thinning effect of the oil out of the engine.

Thus for Maine in the winter season, we need a low viscosity oil and for Florida a high viscosity oil in the summer season.

Therefore, option (c) is correct.

6 0

3 years ago

A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a

almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water ( $\dot m$ ), in kilograms per second, is determined by following formula:

$\dot m = \frac{\eta \cdot \dot W}{g\cdot H}$ (1)

Where:

$\dot W$ - Pump power, in watts.

$\eta$ - Efficiency, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$H$ - Hydrostatic column, in meters.

If we know that $\eta = 0.72$ , $\dot W = 5000\,W$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the mass flow rate of water is:

$\dot m = 14.683\,\frac{kg}{s}$

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump ( $\Delta P$ ), in pascals, is determined by this equation:

$\Delta P = \rho\cdot g\cdot H$ (2)

Where $\rho$ is the density of water, in kilograms per cubic meter.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the pressure difference is:

$\Delta P = 245175\,Pa$

The pressure difference across the pump is 245.175 kilopascals.

4 0

3 years ago

A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co

NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx 32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

$V_{hem}=\frac{2}{3}\pi r^3$

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

$V_{cyl}=\pi r^2\cdot h$

3) Conical bottom

Volume of conical bottom of radius'r' and angle $\theta$ is

$V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}$

Applying the given values we obtain the volume of the container up to cylinder is

$V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}$

Hence the capacity in liters is $V=32.355\times 1000=32355Liters$

3 0

4 years ago

Q9. A cylindrical specimen of a metal alloy 54.8 mm long and 10.8 mm in diameter is stressed in tension. A true stress of 365 MP

wolverine [178]

Answer:

σ = 391.2 MPa

Explanation:

The relation between true stress and true strain is given as:

σ = k εⁿ

where,

σ = true stress = 365 MPa

k = constant

ε = true strain = Change in Length/Original Length

ε = (61.8 - 54.8)/54.8 = 0.128

n = strain hardening exponent = 0.2

Therefore,

365 MPa = K (0.128)^0.2

K = 365 MPa/(0.128)^0.2

k = 550.62 MPa

Now, we have the following data:

σ = true stress = ?

k = constant = 550.62 MPa

ε = true strain = Change in Length/Original Length

ε = (64.7 - 54.8)/54.8 = 0.181

n = strain hardening exponent = 0.2

Therefore,

σ = (550.62 MPa)(0.181)^0.2

7 0

3 years ago