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julsineya [31]
3 years ago
14

1 A hole opened up in the metal being welded is called ____-through.

Engineering
1 answer:
SCORPION-xisa [38]3 years ago
6 0
When a hole opens its called burn through
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Many BLANK apply trial and error to develop a product. Please Help! I have one hour to finish. 30 points
Julli [10]

Answer:

Entrepreneurs?

Explanation:

8 0
3 years ago
Oxides of nitrogen contribute to the formation of?
Nat2105 [25]

Answer:

B. Acid rain.

C. Photochemical smog.

Explanation:

Oxides of nitrogen contribute to the formation of photochemical smog and acid rain. Photochemical smog is a type of smog produced when ultraviolet light from the sun reacts with nitrogen oxides in the atmosphere while on the other hand, when nitrogen oxide react with the water vapor in the atmosphere forming nitric acid which falls on the earth surface with the help of precipitation.

3 0
4 years ago
Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at
andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:  

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that;  distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using  this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0
3 years ago
The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the
soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

                      (dP - y*L*sin(Q))/ L = 2*t / r

                      dP / L - y*sin(Q) = 2*t / r

Where            dP / L = - dP/dx,

                      dP / dx = -2*t / r - y*sin(Q)

Where            r = D /2 ,

                      dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(0)

                      dP / dx = -8 + 0

                      dP/dx   = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

                      dP / dx = 8 - 62.4

                      dP/dx   = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

                      dP / dx = -8 + 62.4

                      dP/dx   = 54.4 lb / ft^3                      

7 0
3 years ago
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