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JulsSmile [24]
3 years ago
8

If ceramic vessels are typed together based as they were all used as storage containers, in spite of the fact that design elemen

ts indicate they are from different time periods, then they have to be
functional typeoperational typesystematic type
Engineering
1 answer:
andriy [413]3 years ago
3 0

If ceramic vessels are typed together based as they were all used as storage containers, in spite of the fact that design elements indicate they are from different time periods, then they have to be  functional typeoperational typesystematic type is given below

Explanation:

The Batiscan site, excavated in the 1960s, produced one of the largest Vinette I collections known to date. Revisiting this ceramic assemblage has revealed more heterogeneity than is generally recognized within the Vinette I type of pottery. Indeed, variations from the typological definition exist, both within and between Early Woodland ceramic collections. A number of diagnostic traits, such as the presence of exterior and interior cord impressions and the absence of decoration, are challenged by the present study. It is hypothesized that part of this variability is chronological, and that the vessels from Batiscan were manufactured closer to the end of the Early Woodland period. However, other factors, such as the frequency and scale of production, and the possible exchange and circulation of ceramic containers, must also be taken into account when interpreting Vinette I variability.l

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D make your car easy to spot in daytime because they do not illuminate a lot
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A sewage lagoon that has a surface area of 10 ha and a depth of 1 m is receiving 8,640 m^3 /d of sewage containing 100 mg/L of b
Marysya12 [62]

Answer: Coefficient= 0.35 per day

Explanation:

To find the bio degradation reaction rate coefficient, we have

k= \frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}

Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

C out= 20 mg/L

V= 10 ha* 1* 10

V= 10⁵ m³

So, k= \frac{8640*100-8640*20}{20*10^5}

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6 0
3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
4 years ago
A hypothetical metal alloy has a grain diameter of 2.4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diamete
Alex

Answer:

The time required is 10.078 hours or 605 min

Explanation:

The formula to apply here is ;

K=(d²-d²₀ )/t

where t is time in hours

d is grain diameter to be achieved after heating in mm

d₀ is the grain diameter before heating in mm

Given

d=5.5 × 10^-2 mm

d₀=2.4 × 10^-2 mm

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t₂=?

n=2.2

First find K

K=(d²-d²₀ )/t₁

K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3

K=(0.051²-0.024²) ÷25/2

K=0.000243 mm²/h

Re-arrange equation for K ,to get the equation for d as;

d=√(d₀²+ Kt)  where now t=t₂

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4 0
3 years ago
A mass of 5 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C.
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Answer: you can watch a video on how to solve this question on you tube

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