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Korvikt [17]
3 years ago
13

What is a business cycle? a period of economic growth followed by economic contraction the amount of time it takes a business to

produce its products a dangerous time for all businesses a period of increased economic growth
Engineering
2 answers:
Rzqust [24]3 years ago
7 0

Business cycle and its growth followed by economic contraction the amount of time it takes a business to produce products in the following way.

Explanation:

The business cycle is the periodic but irregular up-and-down movement in economic activity, measured by fluctuations in real gross domestic product (GDP) and other macroeconomic variables.

A business cycle is typically characterized by four phases—recession, recovery, growth, and decline—that repeat themselves over time.

Economists note, however, that complete business cycles vary in length. The duration of business cycles can be anywhere from about two to twelve years, with most cycles averaging six years in length.

FACTORS THAT SHAPE BUSINESS CYCLES

Volatility of Investment Spending

  • Variations in investment spending is one of the important factors in business cycles. Investment spending is considered the most volatile component of the aggregate or total demand (it varies much more from year to year than the largest component of the aggregate demand, the consumption spending), and empirical studies by economists have revealed that the volatility of the investment component is an important factor in explaining business cycles in the United States.

Momentum

Technological Innovations

Variations in Inventories

Fluctuations in Government Spending

Politically Generated Business Cycles

Monetary Policies

Fluctuations in Exports and Imports

tatyana61 [14]3 years ago
4 0

Answer:

The business cycle, also known as the economic cycle or trade cycle, is the downward and upward movement of gross domestic product around its long-term growth trend. The length of a business cycle is the period of time containing a single boom and contraction in sequence.

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The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphe
Sonbull [250]

Answer:

a. The time required for the tank to empty halfway is presented as follows;

t_1   =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)

b. The time it takes for the tank to empty the remaining half is presented as follows;

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The total time 't', is presented as follows;

t =  \sqrt{2}  \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}

\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} =  \dfrac{v_2}{2 \cdot g}

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh

The time for the tank to drop halfway is given as follows;

\int\limits^{t_1}_0 {} \,  dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh

t_1  =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)

t_1   = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) =  { \dfrac{\sqrt{2}  \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)

t_1   =   { \dfrac{\sqrt{2}  \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)The time required for the tank to empty halfway, t₁, is given as follows;

t_1   =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

\int\limits^{t_2}_0 {} \,  dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh

t_2  =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The total time, t, to empty the tank is given as follows;

t = t_1 + t_2 =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} } =  \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}

t =  \sqrt{2}  \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }

3 0
2 years ago
10. When an adhesion bond is made by melting a filler metal and allowing it to spread into the pores of the
TEA [102]
I believe the correct answer is “A”
8 0
3 years ago
Please help me with this question
Nat2105 [25]

Answer:

The answer is c and the teacher helped me

Explanation:

i had help from the tescger and the assignment is done

5 0
1 year ago
Write a Nested While Loop that will increment the '*' from 1 to 10.
andrey2020 [161]

Answer:

The program is as follows:

i = 1

while(i<11):

   j = 1

   while(j<=i):

       print('*', end = '')

       j += 1

   i += 1

   print()

Explanation:

Initialize i to 1

i = 1

The outer loop is repeated as long as i is less than 11

while(i<11):

Initialize j to 1

   j = 1

The inner loop is repeated as long as j is less than or equal i

   while(j<=i):

This prints a *

       print('*', end = '')

This increments j and ends the inner loop

       j += 1

This increments i

   i += 1

This prints a blank and ends the inner loop

   print()

8 0
3 years ago
A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

3 0
2 years ago
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