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3 years ago

2. What are the separate parts that make up an oxyacetylene flame?

Engineering

1 answer:

amm18123 years ago

6 0

Answer:

Explanation:

Primary combustion where carbon monoxide (CO) and free hydrogen (H) are produced, and the secondary combustion where water vapor (H2O) and carbon dioxide (CO2) are formed.

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The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03

kykrilka [37]

Answer:

Answer is c 0.500

Explanation:

$SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}$

We know that $\rho_{water}=62.42lb/ft^{3}$

Applying values we get

$SpecificGravity=\frac{31.2}{62.4}=0.5$

7 0

3 years ago

A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [1 1 1] direction. Wha

Ivenika [448]

Answer: required tensile stress is 0.889 MPa

Explanation:

Given that;

tensile load is oriented along the [1 1 1] direction

shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane

first we determine

λ which is Angle between [1 1 1] and [1 0 1]

cosλ = [ 1(1) + 1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]

= 2 / √3√2 = 2/√6

Next, we determine ∅ which is angle between [1 1 1] and [1 1 -1]

so,

cos∅ = [ 1(1) + 1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]

cos∅ = [ 2-1] / [√3√3 ]

cos∅ = 1/3

Now, we know that;

σ = T_stress/cosλcosθ

so we substitute

σ = 0.242 / ( 2/√6 × 1/3 )

σ = 0.242 / 0.2721

σ = 0.889 MPa

Therefore, required tensile stress is 0.889 MPa

3 0

3 years ago

The two boxcars A and B have a weight of 20000lb and 30000lb respectively. If they coast freely down the incline when the brakes

kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value

5 0

3 years ago

14. An engine is brought into the shop with a

Lostsunrise [7]

Answer:

B. To accurately measure spark advance, use a timing light that incorporates an

ignition advance meter. The spark advance cannot be determined by listening to the way the engine sounds.

8 0

2 years ago

A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress = 50 MPa

amplitude stress = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first maximum stress and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

$\sigma _{m}$ = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2

50 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........1

and

225 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........2

from eqution 1 and 2 we get maximum and minimum stress

$\sigma _{maximum}$ = 275 MPa ............3

and $\sigma _{minimum}$ = -175 MPa ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio = $\sigma _{minimum}$ / $\sigma _{maximum}$

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = $\sigma _{maximum}$ - $\sigma _{minimum}$

magnitude = 275 - (-175) = 450 MPa

3 0

3 years ago