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s344n2d4d5 [400]
3 years ago
10

_______ is a material property that pertains to local resistance to plastic deformation, such as scratching or denting. It is of

ten measured by poking a material with a very hard tip (like diamond) and measuring the size and depth of the resulting crater that forms.
Engineering
1 answer:
Readme [11.4K]3 years ago
7 0

Answer: hardness

Explanation:

Hardness is a measure of a material's ability to resist plastic deformation. In other words, it is a measure of how resistant material is to denting or scratching. Diamond, for example, is a very hard material. It is extremely difficult to dent or scratch a diamond. In contrast, it is very easy to scratch or dent most plastics.

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A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed
yan [13]

Answer:

b. 1232.08 km/hr

c. 1.02 kn

Explanation:

a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.

See attachment for the remaining.

3 0
3 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
Rainbow [258]

Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6.  The bending fatigue could be handled with specific load requirements  for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the  maximum stress, as in torsional fatigue, which can be performed on  axial-type specially designed machines also, using the proper fixtures if  the maximum twist required is small, in which linear motion is changed to rotational motion.

7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0
3 years ago
Maximum iorsional shear siress.? Select one: a)- occurs at the center of a shaft. b)- occurs at the outer surface of a shaft c)-
pickupchik [31]

Answer:

b). Occurs at the outer surface of the shaft

Explanation:

We know from shear stress and torque relationship, we know that

\frac{T}{J}= \frac{\tau }{r}

where, T = torque

            J = polar moment of inertia of shaft

            τ = torsional shear stress

             r = raduis of the shaft

Therefore from the above relation we see that

            \tau = \frac{T.r}{J}

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.

When r= 0, then τ = 0

and when r = R , τ is maximum

Thus, torsional shear stress is maximum at the outer surface of the shaft.

4 0
3 years ago
A. Calculate the concentration of a toxic (Cout) that is added to a lake with an input concentration of 100 μg/l. The flow rate
Mazyrski [523]

Answer:

Calt =42.129 mg/eit

Explanation:

See attached image

5 0
3 years ago
You buy a 75-W lightbulb in Europe, where electricity is delivered to homes at 240V. If you use the lightbulb in the United Stat
givi [52]

Answer:

The bulb will be \frac{1}{4} times as bright as it is in Europe.

Explanation:

Data provided in the question:

Power of bulb in Europe = 75 W

Voltage provided in Europe = 240 V

Voltage provided in United Stated = 120 V

Now,

We know,

Power = Voltage² ÷ Resistance

Therefore,

75 = 240² ÷ Resistance

or

Resistance = 240² ÷ 75

or

Resistance = 768 Ω

Therefore,

Power in United States = Voltage² ÷ Resistance

= 120² ÷ 768

= 18.75 W

Therefore,

Ratio of powers

\frac{\text{Power in united states}}{\text{Power in Europe}}=\frac{18.75}{75}

= \frac{1}{4}

Hence,

The bulb will be \frac{1}{4} times as bright as it is in Europe.

7 0
3 years ago
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