Uestion 10

Electricians will sometimes call _____ “disconnects” or a “disconnecting means”.

AleksandrR [38]

Answer:
D) Three-pronged plug

Hope it helps u!

4 0

3 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

Air enters a 28-cm diameter pipe steadily at 200 kPaand 208C with a velocity of 5 m/s. Air is heated as it flows, and leaves the

Alina [70]

Answer:

(a) $\dot V_1$ = 0.308 m³/s

(b) $\dot m$ = 0.732 kg/m³

(c) v₂ = 5.94 m/s.

Explanation:

(a) The volume flow rate is given by the cross sectional area of the pipe × Velocity of flow of air

Diameter of pipe = 28 cm = 0.28 m

The cross sectional area, A, of the pipe = 0.28²/4×π = 0.0616 m²

Volume flow rate = 5 × 0.0616 = 0.308 m³/s

$\dot V_1$ = 0.308 m³/s

(b) From the general gas equation, we have;

p₁v₁ = RT₁ which gives;

p₁/ρ₁ = RT₁

ρ₁ = p₁/(RT₁)

Where:

ρ₁ = Density of the air

p₁ = 200 kPa

T₁ = 20 C =

R = 0.287 kPa·m³/(kg·K)

ρ₁ = 200/(0.287 ×293.15) = 2.377 kg/m³

The mass flow rate = Volume flow rate × Density

The mass flow rate, $\dot m$ = 2.377×0.308 = 0.732 kg/m³

$\dot m$ = 0.732 kg/m³

(c) The density at exit, ρ₂, is found from the the universal gas equation as follows;

ρ₂ = p₂/(RT₂)

Where:

p₂ = Pressure at exit = 180 kPa

T₂ = Exit temperature = 40°C = 273.15 + 40 = 313.15 K

∴ ρ₂ = 180/(0.287×313.15) = 2.003 kg/m³

$\dot m$ = ρ₂× $\dot V_2$

$\dot V_2$ = $\dot m$ /ρ₂ = 0.732/2.003 = 0.366 m³/s

$\dot V_2$ = v₂ × A

v₂ = $\dot V_2$ /A = 0.366/0.0616 = 5.94 m/s.

v₂ = 5.94 m/s.

5 0

3 years ago

The connection is made using a bolt and nut and two washers. If the allowable bearing stress of the washers on the boards is (sb

Yuki888 [10]

Answer:

P = 0.490 kip

Explanation:

given data

allowable bearing stress = 2 ksi

allowable tensile stress = 18 ksi

diameter = 0.31 in

outer diameter = 0.75 in

inner diameter (hole) = 0.50 in

solution

we find here cross section area of shank that is express as

Area = $\frac{\pi }{4} \times d^2$ ..................1

area = $\frac{\pi }{4} \times 0.31^2$

area = 0.0754 in²

and

now we get here allowable load in bolt will be

$\sigma = \frac{P}{A}$ ...................2

P = $\sigma \times A$

P = 18 × 10³ × 0.0754

P = 1357.2 = 1.357 kip

and

now find here area of washer is

Area = $\frac{\pi }{4} \times (d^2-d1^2)$ .......................3

put here value

Area = $\frac{\pi }{4} \times (0.75^2-0.5^2)$

area = 0.2454 in²

so now we get here allowable load of washer will be

$\sigma = \frac{P}{Area}$ .....................4

P = 2 × 10³ × 0.245

P = 490 = 0.490 kip

6 0

3 years ago

The fan pressure differential gage on an air handler reads 12 cm H2O. What is this pressure differential in kiloPascals

Deffense [45]

Answer:

$1.18\ \text{kPa}$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

h = Height of reading = 12 cm

$\rho$ = Density of water = $1000\ \text{kg/m}^3$

Pressure due to height difference is given by

$P=\rho gh\\\Rightarrow P=1000\times 9.81\times 12\times 10^{-2}\\\Rightarrow P=1177.2\ \text{Pa}=1.1772\ \text{kPa}\approx 1.18\ \text{kPa}$

The pressure differential is $1.18\ \text{kPa}$ .

4 0

3 years ago