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ASHA 777 [7]
2 years ago
11

If every one-point change in the federal funds rate alters aggregate demand by $200 billion, how far did AD shift in response to

the interest rate cuts?
Physics
1 answer:
attashe74 [19]2 years ago
7 0

If every one-point change in the federal funds rate alters aggregate demand by $200 billion, the AD shifted to the right by a billion. This is further explained below.

<h3>What is the interest rate?</h3>

Generally, the interest rate is simply defined as the part of a loan levied to the borrower as interest, usually represented as a yearly percentage of the loan outstanding

In conclusion, If a one-point increase in the federal funds rate changes aggregates demand by $200 billion, the AD shifts to the right by a billion.

Read more about interest rate

brainly.com/question/13427971

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sveta [45]
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C2H4O2 is what ??????
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An ice skater spins at 2.5 rev/s when his arms are extended. He draws his arms in and spins at 10.0 rev/s. By what factor does h
Rainbow [258]

Answer:

The moment of inertia decreased by a factor of 4

Explanation:

Given;

initial angular velocity of the ice skater, ω₁ = 2.5 rev/s

final angular velocity of the  ice skater, ω₂ = 10.0 rev/s

During this process we assume that angular momentum is conserved;

I₁ω₁ = I₂ω₂

Where;

I₁ is the initial moment of inertia

I₂ is the final moment of inertia

I_2 = \frac{I_1 \omega_1}{\omega_2} = \frac{I_1*2.5}{10} \\\\I_2 = 0.25I_1 = \frac{1}{4}I_1

Therefore, the moment of inertia decreased by a factor of 4

4 0
3 years ago
A coin mass 0.005 kg is dropped from a height of 3 m. How much mechanical energy does it have right before it hits the ground ?
lesya692 [45]
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6 0
3 years ago
If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done
ioda

Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

Given;

power exerted by the lawnmower engine, P = 19 hp

time in which the power was exerted, t = 1 minute = 60 s.

1 hp = 745.7 watts

The work done by the lawnmower is calculated as follows;

Work = Energy = \frac{Power}{time} \\\\Work = \frac{(19 \times 745.7)}{60} = 236.14 \ J

Therefore, the work done by the lawnmower is 236.14 J.

6 0
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