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3 years ago

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If the Central Bank of Macroland puts an additional 1,000 dollars of currency into the economy, the public deposits all currency

into the banking system, and banks have a desired reserve/deposit ratio of 0.10, then the banks will eventually make new loans totaling ______ and the money supply will increase by _______.
a. 9,000; 10,000
b. 9,000; 9,000
c. 10,000; 9,000
d. 10,000; 10,000

1 answer:

FromTheMoon [43]3 years ago

4 0

Answer:

a. 9,000; 10,000

Explanation:

The computation is shown below:

The money multiplier is

= 1 ÷ 0.10

= 10

Now If $1,000 are deposited in banks and the expected reserve ratio is 0.10 ration so the lending amount is $900.

And now if we considered the money multiplier, so it would be increased by

= $900 × $10

= $9,000

And, the increase in money supply is

= $9,000 + $1,000

= $10,000

Hence, the correct option is a.

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Answer:

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A body in the solar system has a period of 10,759.22 days and a perihelion speed of 10.18 km/s. a. Calculate the aphelion radius

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Answer:

Explanation:

From the information given, by applying Kepler's 3rd law,

$T^2 \alpha a^3$

where;

T = period

a = semi major axis

T = 356 days (for earth)

a = 1 AU = 1.496 $\times 10^8 \ km$

Therefore, $T^2 = ca^3$

$c= \dfrac{365^2}{(1.496 \times 10^8)^3}$

c = 3.9791 $\times 10^{20} \ day^2/km^3$

However, if the body in the solar system has a period of 10.759.22 days, then, a =?

∴

$T^2 = ca^3$

$a3 = \dfrac{10759.22^2}{3.9791 \times 10^{-20}}$

$a^3 = 2.9092 \times 10^{27}$

$a= \sqrt[3]{2.9092 \times 10^{27}}$

a = 1.4275 $\times 10^9 \ km$

However, the velocity for a perihelion = 10.18 km/s

Using the formula

$v = \sqrt{GM ( \dfrac{2}{r}-\dfrac{1}{a})}$ to calculate the radius, we have:

G = $6.674 \times 10^{-11}$

M = $1.989\times 10^{30} \ kg$

r = perihelion

$v ^2= GM ( \dfrac{2}{r}-\dfrac{1}{a})$

$(10.18 \times 10^3) ^2= 6.674 \times 10^{-11} \times 1.989 \times 10^{30} ( \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}})$

$7.8068 \times 10^{-13}= \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}}$

$\dfrac{2}{r} = 1.4824 \times 10^{-12}$

$r = \dfrac{2}{1.4824 \times 10^{-12}}$

$r = 1.349 \times 10^{12}$

Similarly, the perihelion is expressed by the equation,

r = a(1 - e)

where ;

e= eccentricity

∴

$1.349 \times 10^{12} = 1.425 \times 10^{12} ( 1 - e)$

$1.349 \times 10^{12} - 1.425 \times 10^{12}= - 1.425 \times 10^{12} (e)$

$-7.6\times 10^{10}= - 1.425 \times 10^{12} (e)$

$\dfrac{-7.6\times 10^{10}}{- 1.425 \times 10^{12}}= (e)$

e ( eccentricity) = 0.0533

Aphelion radius in natural miles, r = a( 1+ e)

$r = 1.425 \times 10^{12} ( 1 + 0.0533)$

$r = 1.50 \times 10^{12} \ m$

to nautical miles, we have:

$r = 1.50 \times 10^{12} \times 0.00054 \ nautical \ mile$

radius of aphelion $\mathbf{r = 8.10 \times 10^8}$ nautical miles

In respect to the value of a( i.e $1.4275 \times 10^9 \ km)$

the body of the solar system is Saturn

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