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2 years ago

Steam locomotives with a 4-6-2 wheel arrangement were usually classified as what?.

Engineering

1 answer:

Korvikt [17]2 years ago

7 0

Answer:

See Explanation

Explanation:

Locomotives with 4 leading wheels, 6 driving wheels and 2 trailing wheels knowns as "Pacific" type locomotives.

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This manometer is used to measure the difference in water level between the two tanks.

SpyIntel [72]

Answer:

a) True

Explanation:

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3 0

2 years ago

Where is the Volkswagen super factory located? how is this locate relevant to us?

Blizzard [7]

It’s in Wolfsburg Germany

4 0

3 years ago

Wave flow of an incompressible fluid into a solid surface follows a sinusoidal pattern. Flow is two-dimensional with the x-axis

Artyom0805 [142]

Answer:

sorry , for my point

Explanation:

4 0

2 years ago

A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by

gregori [183]

Answer:

(a) T = W/2(1-tanθ) (b) 39.81°

Explanation:

(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:

Summation of moment in clockwise direction is equivalent to zero. Therefore,

T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0

T*l*(cosθ - sinθ) = W*(l/2)*cosθ

T = W*cosθ/2(cosθ - sinθ)

Dividing both the numerator and denominator by cosθ, we have:

T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)

(b) If T = 3W, then:

3W = W/2(1-tanθ),

Further simplification and rearrangement lead to:

1 - tanθ = 1/6

tanθ = 1 - (1/6) = 5/6

θ = tan^(-1) 5/6 = 39.81°

8 0

2 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4

Vladimir [108]

Answer:

maximum stress is 2872.28 MPa

Explanation:

given data

radius of curvature = 3 × $10^{-4}$ mm

crack length = 5.5 × $10^{-2}$ mm

tensile stress = 150 MPa

to find out

maximum stress

solution

we know that maximum stress formula that is express as

$\sigma m = 2 ( \sigma o ) \sqrt{\frac{a}{\delta t}}$ ......................1

here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack

so put here all value in equation 1 we get

$\sigma m = 2 ( \sigma o) \sqrt{\frac{a}{\delta t}}$

$\sigma m = 2(150) \sqrt{ \frac{\frac{5.5*10^{-2}}{2}}{3*10^{-4}}}$

σm = 2872.28 MPa

so maximum stress is 2872.28 MPa

8 0

2 years ago