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3 years ago

6

Policy makers in the U.S. government have long tried to write laws that encourage growth in per capita real GDP. These laws typi

cally do one of three things: a. They encourage firms to invest more in research and development in order to boost technology. b. They encourage individuals to save more in order to boost the physical capital stock. c. They encourage individuals to invest more in education in order to boost the stock of human capital. For each of the above three points, please name a law or government program with that intention.

1 answer:

ikadub [295]3 years ago

8 0

Answer:

(a)They encourage firms to invest more in research and development in order to boost technology.

Some of the laws or government programs developed to encourage it include:

Granting patents for inventions
Issuing grants from the National Science Foundation
Issuing grants from the National Institute of Health
Offering tax credits for R&D costs

(b) <u>They encourage individuals to save more in order to boost the physical capital stock</u>

Lowering tax rates on incomes earned through savings
Permitting citizens to open health and savings account
Offering retirement savings plan
Coming up with 529 education savings account

(c) <u>They encourage individuals to invest more in education in order to boost the stock of human capital</u>

Establishment of a system of public schools and universities
Government-provided student loans
Provision of education subsidies for veterans

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3 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

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Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

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Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

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