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3 years ago

As a general rule of thumb, in-line engines are easier to work on than the other cylinder arrangements.

Engineering

1 answer:

siniylev [52]3 years ago

5 0

Answer:

The general rule of thumb is that the SMALLER a substance's atoms and the STRONGER the bonds, the harder the substance. Two of the strongest forms of chemical bonds are the ionic and covalent bonds.

Explanation:

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Determine whether or not each of the following four transaction execution histories is serializable. If a history is serializabl

ludmilkaskok [199]

Answer:

Option D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]

Explanation:

The execution in the option D is correct. This is because there is more than one reasonable criterion.

8 0

3 years ago

A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).

Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress $\sigma_m = 90MPa$

stress amplitude $\sigma_a = 190MPa$

a) $\sigma_m =\frac{\sigma_max+\sigma_min}{2}$

$90 =\frac{\sigma_{max}+\sigma_{min}}{2}$ --------------1

$\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}$

$190 = \frac{\sigma_{max}-\sigma_{min}}{2}$ -----------2

solving 1 and 2 equation we get

$\sigma_{max} = 280MPa$

b) $\sigma_{min} = - 100MPa$

stress ratio $=\frac{\sigma_{min}}{\sigma_{max}}$

$=\frac{-100}{280} = -0.35$

d)magnitude of stress range

$=(\sigma_{max} -\sigma_{min})$

= 280 -(-100) = 380 MPa

3 0

3 years ago

Which of the following Identifies the challenges faced by scientists experimenting with using artificial photosynthesis as

bezimeni [28]

Answer:

The cost and size of materials needed to produce energy

Explanation:

Artificial photosynthesis is a chemical process that uses solar cells instead of chlorophyll to absorb sunlight and convert it into electricity. This process uses artificial leaves that require man-made catalyst to spilt water present in the air into hydrogen and oxygen. It is clear that the reaction requires heat from the sun for energy production thus the technology is expensive to be applied in most areas of the world. Additionally, results obtained from previous undertaken projects of this type has been ineffective and unsustainable because it involves a lot of trial and error.

8 0

3 years ago

A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force o

Rus_ich [418]

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

8 0

3 years ago

A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The channel undergoes a smooth, gradu

Trava [24]

Answer:

Depth in the contracted section = 2.896m

Velocity in the contracted section = 2.072m/s

Explanation:

Please see that attachment for the solving.

Assumptions:

1. Negligible head losses

2. Horizontal channel bottom

7 0

3 years ago