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3 years ago

What is engineering?

Engineering

1 answer:

anzhelika [568]3 years ago

3 0

Answer:

Engineering is the use of science and math to design or make things. People who do engineering are called engineers. They learn engineering at a college or university. Engineers usually design or build things. Some engineers also use their skills to solve technical problems.

Explanation:

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You are preparing to exit from an expressway. You should begin slowing to the posted safe

wlad13 [49]

Answer:

Your answer will be B: At least 100 feet after leaving the expressway

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3 years ago

Read 2 more answers

Given in the following v(t) signal.

tatuchka [14]

Answer:

Check the v(t) signal referred to in the question and the solution to each part in the files attached

Explanation:

The detailed solutions of parts a to d are clearly expressed in the second file attached.

3 0

3 years ago

A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co

NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx 32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

$V_{hem}=\frac{2}{3}\pi r^3$

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

$V_{cyl}=\pi r^2\cdot h$

3) Conical bottom

Volume of conical bottom of radius'r' and angle $\theta$ is

$V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}$

Applying the given values we obtain the volume of the container up to cylinder is

$V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}$

Hence the capacity in liters is $V=32.355\times 1000=32355Liters$

3 0

3 years ago

An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a speed of 10 ft/s thr

Sladkaya [172]

Answer:

Part A

The mass plow rate, is approximately 97.0 lbm/s

Part B

The power used to overcome friction, is approximately 1.9 hp

Explanation:

The efficiency of the pump, η = 80%

The power input to the pump, P = 20 hp

The speed of the water through the pipe, v = 10 ft./s

The diameter of the pipe, d = 5.2 inches = 13/30 ft.

The free surface of the pool above the lake, h = 80 ft.

The density of the water, ρ = 62.4 lbm/ft.³

Part A

The mass plow rate, $\dot m$ = Q × ρ

Where;

ρ = 62.4 lbm/ft³

Q = A × v

A = The cross-sectional area of the pipe

∴ Q = π·d²/4 × v = π × ((13/30 ft.)²)/4 × 10 ft.s ≈ 1.4748 ft.³/s

∴ The mass plow rate, $\dot m$ ≈ 1.4748 ft.³/s × 62.4 lbm/ft.³ = 97.02752 lbm/s

The mass plow rate, $\dot m$ ≈ 97.0 lbm/s

Part B

The power to pump the water at the given rate, $P_w$ = $\dot m$ ·g·h

∴ $P_w$ = 97.02752 lbm/s × 32.1740 ft./s² × 80 ft. ≈ 14.1130725 Hp

$P_w$ ≈ 14.1130725 Hp

The power output of the pump, $P_{out}$ = 0.8 × 20 hp = 16 hp

Therefore, the power used to overcome friction, $P_f$ = $P_{out}$ - $P_w$

∴ $P_f$ ≈ 16 hp - 14.1130725 Hp ≈ 1.8869275 hp

The power used to overcome friction, $P_f$ ≈ 1.9 hp

7 0

2 years ago

A 8.0-m-diameter parachute of a new design is to be used to transport a load from flight altitude to the ground with an average

S_A_V [24]

Answer:

drag coefficient for the parachute is Cd = 1.84177

Explanation:

given data

diameter = 8.0 m

average vertical speed = 3 m/s

total weight of load = 200 N

to find out

drag coefficient for the parachute

solution

we will apply here drag coefficient formula that is express as here

drag coefficient Cd = $\frac{2w}{\rho v^2A}$ ......................1

here w is total load and A is area and v is speed

and here ρ = 1.22 kg/cu

put here value we get

Cd = $\frac{2*4*200}{1.229*5^2*\pi * 3^2}$

Cd = 1.84177

drag coefficient for the parachute is Cd = 1.84177

3 0

2 years ago