A socket can be driven using any of the following except for (A) a socket ratchet

Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 2, determine the magnitude of the load

IceJOKER [234]

Answer: 50

Explanation:

7 0

3 years ago

The process by which natural forces move weathered rock and soil from one place to another is called

melomori [17]

Answer:

Erosion is the geological process in which earthen materials are worn away and transported by natural forces such as wind or water. A similar process, weathering, breaks down or dissolves rock, but does not involve movement.Erosion is the opposite of deposition, the geological process in which earthen materials are deposited, or built up, on a landform.

Most erosion is performed by liquid water, wind, or ice (usually in the form of a glacier). If the wind is dusty, or water or glacial ice is muddy, erosion is taking place. The brown color indicates that bits of rock and soil are suspended in the fluid (air or water) and being transported from one place to another. This transported material is called sediment.

Explanation:

Physical erosion describes the process of rocks changing their physical properties without changing their basic chemical composition. Physical erosion often causes rocks to get smaller or smoother. Rocks eroded through physical erosion often form clastic sediments. Clastic sediments are composed of fragments of older rocks that have been transported from their place of origin.Landslides and other forms of mass wasting are associated with physical weathering. These processes cause rocks to dislodge from hillsides and crumble as they tumble down a slope. 

Landslides and other forms of mass wasting are associated with physical weathering. These processes cause rocks to dislodge from hillsides and crumble as they tumble down a slope. 

Plant growth can also contribute to physical erosion in a process called bioerosion. Plants break up earthen materials as they take root, and can create cracks and crevices in rocks they encounter.

<h2>I HOPE I HELPED YOU</h2>

6 0

3 years ago

Read 2 more answers

Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manome

oksano4ka [1.4K]

Answer:

a) Δh = 2 cm, b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1

y₁ = y₂

subtitute

P₁ = P₂ + ½ ρ v₂²

P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

ΔP =ρ_{Hg} g h - ρ g h

ΔP = (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

ΔP = P₁ - P₂

(ρ_{Hg} - ρ) g h = ½ ρ v²

v = $\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}$

in this expression the densities are constant

v = A √h

A = $\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }$

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

A = $\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }$

A = 454.55

we substitute

v = 454.55 √h

to calculate the uncertainty or error of the velocity

h = $\frac{1}{454.55^2} \ v^2$

Δh = $\frac{dh}{dv}$ Δv

$\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}$

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

$\frac{\delta v}{v} = 0.05$

$\frac{\Delta h}{h}$ = 2 0.05

Δh = 0.1 h

Δh = 0.1 20 cm

Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

$\frac{\Delta h}{h}$ = 2 0.01

Δh = 0.02 h

Δh = 0.02 20

Δh = 0.1 20 cm

Δh = 0.4 cm = 4 mm

5 0

3 years ago

An oil reservoir has a average porosity of 20%, an area of 100 acres, and a av- erage thickness of 10 feet. The connate water sa

hram777 [196]

Answer:a) recovery factor = 53.3% ,bi) volume of oil per acre-ft = 11090.64bbl/acre-ft

bii) bulk volume of the reservoir in acre-ft = 1000 acre-ft

c) 62214.74 cubic feet

Explanation: a) recovery factor is the percentage amount of oil that can recovered from a reservoir, it is the oil produced divided by oil initially in place

Recovery factor= 1 - (Soi/1-Swi)

= 1 - (0.35/1-0.25)

= 0.533 × 100%

= 53.3%

bi) volume of oil which may be recovered per acre-ft = 7758.porosity.(1- Swi-Soi)

= 7758 x 0.2 x (1-0.25-0.35)

= 620.64 bbl/acre-ft

bii) bulk volume of the reservoir in acre-ft= Area x thickness

= 100 acres x 10 ft

= 1000 acre-ft

c) total volume of oil in cubic feet

since we have gotten volume as 1000 acre-ft we simply multiply it by the volume of oil gotten in answer bi)

= 1000 acre-ft x 620.64 bbl/ acre-ft

= 620640bbl

So we convert from barrel(bbl) to cubic feet, and 1 barrel is equal to 5.609 cubic feet, so to convert the answer from barrel to cubic feet we multiply the answer 620640 bbl by 5.609

= 620640 bbl x 5.609

= 3481580.711 cubic feet

4 0

3 years ago

How should the system administrator resolve the issue?

Hunter-Best [27]

Answer:

Option B

Select the Marketing User checkbox in the user record

Explanation:

The user settings may limit users who can view and set up advanced campaigns. In case there's no keyboard shortcut already programmed to create a new campaign and also the button for creating new campaign isn't visible, the best way to go about it is by selecting the marketing user checkbox in the user record and create a new campaign,

4 0

4 years ago