Answer:
the elevation at point X is 2152.72 ft
Explanation:
given data
elev = 2156.77 ft
BS = 2.67 ft
FS = 6.72 ft
solution
first we get here height of instrument that is
H.I = elev + BS ..............1
put here value
H.I = 2156.77 ft + 2.67 ft
H.I = 2159.44 ft
and
Elevation at point (x) will be
point (x) = H.I - FS .............2
point (x) = 2159.44 ft - 6.72 ft
point (x) = 2152.72 ft
Answer:
δu/δx+δu/δy = 6x-6x =0
9r^2
Explanation:
The flow is obviously two-dimensional, since the stream function depends only on the x and y coordinate. We can find the x and y velocity components by using the following relations:
u =δψ/δy = 3x^2-3y^2
v =-δψ/δx = -6xy
Now, since:
δu/δx+δu/δy = 6x-6x =0
we conclude that this flow satisfies the continuity equation for a 2D incompressible flow. Therefore, the flow is indeed a two-dimensional incompressible one.
The magnitude of velocity is given by:
|V| = u^2+v^2
=(3x^2-3y^2)^2+(-6xy)^2
=9x^4+18x^2y^2+9y^2
=(3x^2+3y^2)^2
=9r^2
where r is the distance from the origin of the coordinates, and we have used that r^2 = x^2 + y^2.
The streamline ψ = 2 is given by the following equation:
3x^2y — y^3 = 2,
which is most easily plotted by solving it for x:
x =±√2-y^3/y
Plot of the streamline is given in the graph below.
Explanation for the plot: the two x(y) functions (with minus and plus signs) given in the equation above were plotted as functions of y, after which the graph was rotated to obtain a standard coordinate diagram. The "+" and "-" parts are given in different colors, but keep in mind that these are actually "parts" of the same streamline.
Answer:
Chipping; cutting; drilling; grinding
Explanation:
Activities such as abrasive blasting with sand; sawing brick or concrete; sanding or drilling into concrete walls; grinding mortar; manufacturing brick, concrete blocks, stone countertops, or ceramic products; and cutting or crushing stone result in worker exposures to respirable crystalline silica dust.
Answer:
Modal analysis is performed to identify the modes, and the response in that mode can be picked from the response spectrum. These peak responses are then combined to estimate a total response. A typical combination method is the square root of the sum of the squares (SRSS) if the modal frequencies are not close.
Answer:
L= 312.75 mm
Explanation:
given data
elastic modulus E = 106 GPa
cross-sectional diameter d = 3.9 mm
tensile load F = 1660 N
maximum allowable elongation ΔL = 0.41 mm
to find out
maximum length of the specimen before deformation
solution
we will apply here allowable elongation equation that is express as
ΔL =
....................1
put here value and we get L
L = 
solve it we get
L = 0.312752 m
L= 312.75 mm