Answer:
Some of the benefits are tangible for they are visible in the design and production process, while the other benefits are intangible which may not be visible directly but result in improvement in the quality of product, better control over designing and production process, reduction of stress on the designers etc.
Answer:
critical stress required for the propagation is 27.396615 ×
N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 ×
N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) =
.....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 ×
m
so now put value in equation 1 we get
( σc ) =
( σc ) =
( σc ) = 27.396615 ×
N/m²
so critical stress required for the propagation is 27.396615 ×
N/m²
Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm
Required:
•). The magnitude of current= ?
•••••••••••••SOLUTION•••••••••••••
We can find the relation ship between current, voltage and resistance with the help of Ohms law.
According to ohms law;
V= IR.
Rearranging the above equation;
I= V/ R
Putt the values in the above equation; we get
I= 15V/ 1000ohm
I = 0.015 A( ampere)
••••••••••••••• CONCLUSION•••••••
The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.
Answer:
80grit
Explanation:
80 grit is coarsest grit that may be used on aluminum
The lowest grit sizes range from 40 to 60. From the given options 80 grit is practically available grit.
What is a sandpaper used for?
They are essentially used for surface preparation. Sandpaper is produced in a range of grit sizes and is used to remove material from surfaces, either to make them smoother (for example, in painting and wood finishing), to remove a layer of material (such as old paint), or sometimes to make the surface rougher (for example, as a preparation for gluing).
Answer:
The elevation at the high point of the road is 12186.5 in ft.
Explanation:
The automobile weight is 2500 lbf.
The automobile increases its gravitational potential energy in
. It means the mobile has increased its elevation.
The initial elevation is of 5183 ft.
The first step is to convert Btu of potential energy to adequate units to work with data previously presented.
British Thermal Unit -
Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:
Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.
is the final elevation and
is the initial elevation.
Replacing W in the Ep equation
Finally, the next step is to replace the variables of the problem.
The elevation at the high point of the road is 12186.5 in ft.