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aleksandrvk [35]
3 years ago
14

Based on the following passage, why might the government use the U.S. Army Corps of Engineers to undertake hydroelectric power p

rojects instead of allowing private businesses to undertake them?
Before World War I, most power projects were developed by private interests. During the Great Depression, president Franklin Roosevelt developed federal hydropower projects to give low-cost energy to consumers. The U.S. Army Corps of Engineers undertook three major hydroelectric power projects during this era: the Bonneville Dam on the Columbia River, the Fort Peck Dam on the Missouri River, and the Passamaquoddy Tidal Power Project in Maine. Congress organized the Bonneville Power Administration in 1937 to distribute the power and set usage rates for all power generated at Bonneville Dam.


Hydroelectric power is closely linked to national security.

Private businesses are by law disallowed from engaging in hydroelectric power projects.

The government would have less power to control the cost of electricity from privately funded projects.

Private businesses have less expertise in hydroelectric power than the U.S. Army Corps of Engineers.
Engineering
1 answer:
mr Goodwill [35]3 years ago
3 0

Answer:

The government would have less power to control the cost of electricity from privately funded projects.

Explanation:

According to the given excerpt, the government, the government decided to take over most of the hydroelectric power projects as opposed to privately owned corporations which previously held a monopoly. The major reason was to provide low-cost energy to consumers.

This decision taken by the government to use the Army Corps of Engineers to undertake hydroelectric power projects instead of allowing private businesses to undertake them was to allow the government have more power to control the cost of electricity.

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Would be much appreciated if someone could help with this will give brainiest.
Mashcka [7]

Answer:   both mm and inches on each dimension in a sketch (with the main dimension in one format and the other in brackets below it), in the way you can have dual dimensions shown when detailing an idw view.

personally think it would look a mess/cluttered with even more text all over the sketch environment, but everyone's differenent.

If it's any help - you know you can enter dimensions in either format?  If you're working in mm you can still dimension a line and type "2in" and vice-versa.  Probably know this already, but no harm saying it, just in case.

You can enter the units directly in or mm and Inventor will convert to current document settings (which  you can change - maybe someone can come up with a simple toggle icon to toggle the document settings).  Tools>Document Settings>Units

Unlike SolidWorks when you edit the dimension the original entry shows in the dialog box so it makes it easy to keep track of different units even if they aren't always displayed.  (SWx does the conversion or equation and then that is what you get.)

I work quite a bit in inch and metric and combination (ex metric frame motor on inch machine) and it doesn't seem to be a real difficulty to me.

4 0
3 years ago
If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo
Stolb23 [73]

Answer:

526.5 KN

Explanation:

The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.

But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.

h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg

where ρ = density of the fluid and g = acceleration due to gravity

h = ΔP/ρg

ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa

Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with

Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa

Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²

Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN

3 0
4 years ago
2.) A fluid moves in a steady manner between two sections in a flow
Talja [164]

Answer:

250\ \text{lbm/min}

625\ \text{ft/min}

Explanation:

A_1 = Area of section 1 = 10\ \text{ft}^2

V_1 = Velocity of water at section 1 = 100 ft/min

v_1 = Specific volume at section 1 = 4\ \text{ft}^3/\text{lbm}

\rho = Density of fluid = 0.2\ \text{lb/ft}^3

A_2 = Area of section 2 = 2\ \text{ft}^2

Mass flow rate is given by

m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}

The mass flow rate through the pipe is 250\ \text{lbm/min}

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}

The speed at section 2 is 625\ \text{ft/min}.

3 0
3 years ago
What is the primary function of NCEES?
charle [14.2K]

Answer:

It is a non profit organization that dedicates to licensing professional engineers and surveyors

Explanation:

6 0
3 years ago
A large increase in elevation can cause a carbureted engine to run ________ if not properly adjusted for the altitude. a Rich b
mash [69]

Answer:

B - Poor

Explanation:

As you get higher up, There is less oxygen which causes the engine to create less power.

3 0
3 years ago
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