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4 years ago

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When economists determine that a nation’s GDP has declined, they can point to this as a sign of economic shrinkage. economic gro

wth. low unemployment. poor leadership

2 answers:

amid [387]4 years ago

8 0

That's economic shrinkage.

GrogVix [38]4 years ago

4 0

ANSWER: Economic Shrinkage

EXPLANATION: When a country's GDP falls, it affects the whole country negatively. Fall in GDP affects the consumers, businesses and investors. The effect is best seen in the numbers of stock markets and currency exchange rate. If the GDP goes down by even 2%, Foreign Direct Investments (FDI) slow down and interest rates go up. This whole scenario is known as Economic Shrinkage.

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Which of the following most directly shows how society affects physics? A. Elected officials make decisions about what research

Scorpion4ik [409]

The option above which most directly shows how society affects physics is that physicists work on projects that will get them nominated for Nobel prize.

<h3>Who is a physicist?</h3>

A physicist is a person who who studies physics.

Physics is one of the branches of science which involves the study of matter, light, mechanics...

In conclusion: the option below which most directly shows how society affects physics is that physicists work on projects that will get them nominated for Nobel prize.

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2 years ago

How did equity in the 2008 crisis?

MArishka [77]

Answer:

While the causes of the bubble are disputed, the precipitating factor for the Financial Crisis of 2007–2008 was the bursting of the United States housing bubble and the subsequent subprime mortgage crisis, which occurred due to a high default rate and resulting foreclosures of mortgage loans, particularly adjustable. Hope this helps!

Explanation:

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4 years ago

According to car and driver, an alfa romeo going 70 mph requires 177 feet to stop. assuming that the stopping distance is propor

nadezda [96]

Fewhdknhs fdddkp,'sadjofipw]e

7 0

3 years ago

A particle with charge − 3.74 × 10 − 6 C is released at rest in a region of constant, uniform electric field. Assume that gravit

dalvyx [7]

Answer:

57.5022228905 s

Explanation:

$V_i$ = Initial voltage

$V_f$ = Final voltage = 0.401 V

$t_1$ = Initial time = 8.36 s

$t_2$ = Final time

K = Kinetic Energy = $3.18\times 10^{-8}\ J$

C = Charge = $3.74\times 10^{-6}\ C$

Voltage is given by

$V=\dfrac{K}{C}\\\Rightarrow V=\dfrac{3.17\times 10^{-8}}{3.74\times 10^{-6}}\\\Rightarrow V=0.00847593582888\ V$

We have the relation

$\dfrac{V_i}{V_f}=(\dfrac{t_i}{t_f})^2\\\Rightarrow t_f=\sqrt{\dfrac{t_i^2V_f}{V_i}}\\\Rightarrow t_f=\sqrt{\dfrac{8.36^2\times 0.401}{0.00847593582888}}\\\Rightarrow t_f=57.5022228905\ s$

The time taken is 57.5022228905 s

4 0

3 years ago

A ball is thrown from the top of a building with an initial velocity of 21.9 m/s straight upward, at an initial height of 51.6 m

nalin [4]

Part A)

when ball will reach to highest point then it's speed will become zero

so we can use kinematics to find the time

$v_f = v_i + at$

$0 = 21.9 + (-9.8) t$

$0 = 21.9 - 9.8 t$

$t = 2.23 s$

Part b)

for finding the maximum height we can use another kinematics equation

$v_f^2 - v_i^2 = 2ad$

$0 - 21.9^2 = 2(-9.8)(H)$

$H = \frac{21.9^2}{19.6} = 1.12 m$

so it will rise to 1.12 m from the point of projection

Part C)

Ball will take double the time which it take to reach the top point.

So here the time to reach the top is 2.23 s

so time taken by the ball to reach at same point after projection is given as

$t = 2(2.23) = 4.46 s$

Since ball have reached to same point so the final velocity must be same as initial velocity

so we have

$v_f = 21.9 m/s$ downwards

Part d)

when ball reached to the bottom

displacement of ball = -51.6 m

$a = -9.8 m/s^2$

$v_i = 21.9 m/s$

now by kinematics we have

$d = v_i t + \frac{1}{2}at^2$

$-51.6 = (21.9)t + \frac{1}{2}(-9.8)t^2$

$4.9 t^2 - 21.9 t - 51.6 = 0$

by solving above equation we have

$t = 6.2 s$

now for the velocity at that instant we have

$v_f = v_i + at$

$v_f = 21.9 - (9.8) (6.2)$

$v_f = -38.6 m/s$

so its velocity is 38.6 m/s downwards

Part e)

for the position of ball at t = 5.35 s we can use

$d = v_i t + \frac{1}{2}at^2$

$d = 21.9(5.35) + \frac{1}{2}(-9.8)(5.35)^2$

$d = -23.1 m$

so it is 23.1 m below the initial position from which it is thrown

now for the velocity we can say

$v_f = v_i + at$

$v_f = 21.9 + (-9.8)(5.35)$

$v_f = -30.53 m/s$

so it will be moving downwards with speed 30.53 m/s

8 0

3 years ago