A properly fitted wearable pfd should have which characteristics

Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m

Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0

3 years ago

A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f

KIM [24]

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) $\frac{dm}{2}$ ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = $\frac{L}{\pi dm}$ ...............2

here lead L = n × p

so tan(α) = $\frac{2\times 2}{\pi 19}$

α = 3.83°

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) $\frac{19\times 10^{-3}}{2}$

T = 1.59 Nm

so

power = $\frac{2\pi N \ T }{60}$ .................3

put here value

power = $\frac{2\pi \times 300\times 1.59}{60}$

power = 49.95 W

and

as φ > α

so it is self locking screw

8 0

3 years ago

Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of

yKpoI14uk [10]

Answer: Hello the question is incomplete below is the missing part

Question: determine the temperature, in °R, at the exit

answer:

T2= 569.62°R

Explanation:

T1 = 540°R

V2 = 600 ft/s

V1 = 60 ft/s

h1 = 129.0613 ( value gotten from Ideal gas property-air table )

<em>first step : calculate the value of h2 using the equation below </em>

assuming no work is done ( potential energy is ignored )

h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778

∴ h2 = 136.17 Btu/Ibm

From Table A-17

we will apply interpolation

attached below is the remaining part of the solution

8 0

3 years ago

Hydraulic fluid is primarily used to

Lerok [7]

Answer:

hydraulic flute is energy transfer medium in all hydraulic system this simple function is only achieved by flute that does not easily trap gases trapped gas and forming problems should bring a higher level of compressibility to a fluid than its usual at a point to support a very steep pass trekking system

8 0

3 years ago

Write using about 10-15 lines for each of the six materials (metals, ceramics, glasses, polymers, composites, and semiconductors

Svetradugi [14.3K]

Answer:

See Answer below- Explanation is the entire answer

Explanation:

Metals:

Properties: Ductile, good heat conductivity, good electrical conductivity, high strength;

Drawbacks: Relatively high weight, reactive with oxygen to create oxides- corrosion is presented;

Examples: steel, aluminum alloys, brass, copper, titanium

Applications: Body of the vehicles, structures in the skyscrapers, cooking pots.

Ceramics:

Properties: Brittle, poor heat conductors, poor electrical conductors, high wear resistance, corrosion resistance;

Drawbacks: Deforms by fracturing, shock resistance is low, no conductivity of electricity;

Examples: concrete, tungsten carbide, diamond

Applications: bricks for constructions, clay pots to keep heat, cutting tools for metals;

Glasses:

Properties: amorphous, transparent, high weight

Drawbacks: poor conductors of heat and electricity; brittle; low shock resistance;

Examples: Silica, lead glass, glaze;

Applications: windows, protection screens;

Polymers:

Properties: low density, recyclable, poor heat and electrical conductors, plastic deformation;

Drawbacks: low strength, low operating temperatures;

Examples: polyethylene, nylon, ABS-plastic, rubber;

Applications: toys, tires, insulation covers for the wires.

Composites:

Properties: high strength to weight ratio, can get combination of properties from the used materials, rarely conductive, good shock resistance;

Drawbacks: high cost, hard to recycle, expensive;

Examples: steel-reinforced concrete, carbon fiber, fiber glass, Nomex, sandwich roof panels;

Applications: buildings, bullet proof vests, body of the Formula 1 cars, rockets, roof panels.

Semiconductors:

Properties: brittle, change conductive behavior under certain scenario, poor heat conductors;

Drawbacks: hard to manufacture, expensive;

Examples: Silicon-based semiconductors, Germanium-based semiconductors, Ga-based semiconductors;

Applications: chips, LED, diodes, transistors, op-amps, microprocessors.

8 0

3 years ago