Question 5 1 pts Lumber which has been put through a planer is known as surfaced or ]

True or False? Early engineers used a trial-anderror approach, rather than mathematical and scientific principles when solving

yarga [219]

Answer:

<h2>True Most Especially in the field of Automotive Engineering</h2>

Explanation:

Normally, before the introduction of vehicle diagnostics when a vehicle, mostly automobile/car break down, one could be the vehicle mechanic would only suspect one or two related faults based on the present working condition of the car, the mechanic would perform some trial and error before he could fix the car.

But in recent times, the introduction of vehicle diagnostics devices and software has changed the order as vehicles can be connected to a computer that will scan and tell what the problem is before a possible fix.

3 0

3 years ago

Which of the following describes one of an employee's responsibilities under OSHA's rules?

kow [346]

Explanation:

what are the options for this question?

5 0

3 years ago

Read 2 more answers

Activity

Sphinxa [80]

Answer:

yo do you still need an

answer

Explanation:

6 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

What is the correct expression of the internal moment in segment AB? A 10-ft-long cantilever beam is fixed at end A (x = 0). A u

Komok [63]

Answer:

Option A

Question (in proper order)

What is the correct expression of the internal moment in segment AB? A 10-ft-long cantilever beam is fixed at end A (x = 0). A uniform distributed load of 2 kip/ft is applied between A and B (x = 6 ft). A downward 10-kip force is applied at B, and an upward 8-kip force is applied at end C (x = 10 ft). In addition, a clockwise couple moment of 40 kip-ft is exerted at the free end C. What is the correct expression of the internal moment in segment AB?

(First Diagram attached below)

$a. -x^{2} + 14x -56 kip.fit \\b. -x^2 + 14x kip.fit\\c. -x^2\\d. -x^{2} + 14x -24\\e. -x^{2} + 24\\f. -x^{2} - 56$

Explanation:

(From the second diagram attached below)

Note that

Moment of a force is the product of the force and the perpendicular distance

sum of horizontal forces, ∑Fx = 0 (that is sum of forces acting towards left = sum of forces acting towards right)

that is zero in this case

Sum of vertical forces, ∑Fy = 0 ( that is sum of forces acting upward = sum of forces acting downward)

Total upward force = Total downward force

Ay + 8 = 10 + 2 x 6

where Ay is the force acting at the fixed end

Ay + 8 = 10 + 12

Ay = 22 - 8

Ay = 14 Kip ( force is measured in kip while distance is measured in ft)

Sum of clockwise moment = Sum of anticlockwise moment

let us assume a counterclockwise moment about point A = Ma

8 x 10 + Ma = 40(counterclockwise moment) + (2 x 6) x 3(total uniform force assumed to act at the middle of the 6ft length) + 10 x 6

80 + Ma = 40 + 36 + 60

Ma = 136 - 80

Ma = 56 kip-fit

(From the third diagram attached below)

Consider a section of beam AB from A to X and of distance x fit

let the counterclockwise moment about Xx section be Mx

if we take moment about the Xx section

then

summation of moment about the Xx section ∑Mxx = 0

hence,

Mx + 56 + {(2 * x)[total of the uniform force] * x/2[acting at the middle point of the x ft distance]} = Ay * x

Mx + 56 + 2x * x/2 = 14 * x

Mx + 56 + x^2 = 14x

Mx = $-x^2 + 14x - 56$ kip-ft

8 0

3 years ago