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2 years ago

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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A comet is first seen at a distance of d astronomical units from the Sun and it is traveling with a speed of q times the Earth’s

saw5 [17]

To solve this problem it is necessary to take into account the concepts of Gravitational Force and Kinetic Energy.

The kinetic energy is given by the equation:

$F= \frac{mv^2}2$

La energía gravitacional por,

$F=\frac{GM_cm}{d}$

Where m is the mass, v is the velocity, G the gravitational constant $M_e$ the mass of the earth, m the mass of the sun and d the distance ..

The sum of the energies, we must be a total energy

$E= \frac{mv^2}2+\frac{GM_em}{d}$

By the type of orbit we know that

E> 0 is a hyperbolic orbit

E = 0 is a parabolic orbit

E <0 is a closed orbit.

In the case of hyperbolic orbit

E>0

$\frac{mq^2}{2}-\frac{GM_em}{d}>0\\\frac{qv^2_e}{2}>\frac{GM_em}{d}\\q^2d>2\frac{GM_e}{v^2_e}\\q^2d>2$

The case of the comet is a closed orbit, so,

E<0

$\frac{mv^2}2+\frac{GM_em}{d}$

For parabolic orbit

E=0

$\frac{mv^2_eq^2}{2}-\frac{GM_cm}{d}=0\\\frac{v^2_eq^2}{2}=\frac{GM_c}{d}\\q^2d=2\frac{GM_e}{v^2_e}\\q^2d=2$

For the sun and the earth

$\frac{m_ev_e^2}{r}=\frac{GM_em_e}{r^2}$

$v^2_e=\frac{GM_e}{r}\\\frac{GM_e}{v_e}=r$

where $R \approx 1AU$

$q^2d$ For elliptical orbit

8 0

3 years ago

two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters

Free_Kalibri [48]

Answer:

0 m/s²

Explanation:

Since each skater pushes the other with a force of 70 N, according to Newton's third law, there is an equal reaction and thus the other pushes back with a force of 70 N in the other direction, so we have forces of +70N and -70 N respectively. So, the net force on each skateboarder is F = + 70 N + (-70 N) = + 70 N - 70 N = 0 N.

Since force, F = ma where a = acceleration and m = mass,

a = F/m.

So, since for each skater, F = 0N,

a = 0 N/m

= 0 m/s²

So, the acceleration of each skater is 0 m/s²

5 0

3 years ago

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

3 years ago

Pick the correct word to fill up the blank in below question: velocity, combustion, acceleration, gravitation, force

scZoUnD [109]

Answer: Acceleration

4 0

3 years ago

Which best describes the electric field created by a positive charge?

Vlad [161]

Its rays point away from the charge

3 0

3 years ago

Read 2 more answers