which view of the macroeconomy suggests that the speed of adjustment Purcell correction would be very quick

Physics

1 answer:

horsena [70]2 years ago

7 0

Rational expectations theory suggests that the speed of adjustment Purcell correction would be very quick.

<h3>What Is Rational Expectations Theory?</h3>

The rational expectations theory is a widely used concept and modeling technique in macroeconomics. Individuals make decisions based on three primary factors, according to the theory: their human rationality, the information available to them, and their past experiences.

The rational expectations hypothesis was originally suggested by John (Jack) Muth 1 (1961) to explain how the outcome of a given economic phenomena depends to a certain degree on what agents expect to happen.

People who have rational expectations always learn from their mistakes.
Forecasts are unbiased, and people make decisions based on all available information and economic theories.
People understand how the economy works and how government policies affect macroeconomic variables like the price level, unemployment rate, and aggregate output.

To learn more about Rational expectations theory from the given link

brainly.com/question/16479910

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The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
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where
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$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
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If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

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