All categories

2 years ago

All else equal, the payback period for a project will decrease whenever the_______.

Physics

2 answers:

telo118 [61]2 years ago

3 0

Answer:

(B) cash inflows are moved earlier in time.

Explanation:

The payback period stated time-frame during which the initial amount of investment should be recovered. It is expressed in the year form

The formula to compute the payback period is shown below:

Payback period = Initial investment ÷ Net cash flow

where,

The net cash flow = annual net operating income + depreciation expenses

The payback period of the project decreases when the accumulated starting year cash flows increases that results the movement of the cash inflows earlier in time

just olya [345]2 years ago

3 0

Answer:

Explanation:

because the project is increases

You might be interested in

5. What is the density of 4.5 mL of a liquid that has a mass of 1.3 grams?

antoniya [11.8K]

Answer:

A. 0.289g/mL

Explanation:

Using the equation for density which is d = m/v or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.

5 0

2 years ago

Read 2 more answers

A student drops two metallic objects into a 120-g steel con- tainer holding 150 g of water at 25°C. One object is a 200-g cube o

marissa [1.9K]

Answer:

The mass of the aluminum chunk is 258 g

Explanation:

Given;

mass of steel container = 120-g

mass of water = 150 g

initial temperature of water, = 25°C

mass of copper cube, $M_{cu}$ = 200 g

initial temperature of the copper cube, $T_c_u$ = 85°C

initial temperature of the aluminum chunk $T_A_l$ = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

$M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}$

where;

$C_{AL}$ is specific heat capacity of aluminum

$\delta T_{Al}$ is change in temperature of aluminum

$C_c_u$ is the specific heat capacity of copper

$\delta T_c_u$ is the change in temperature of copper

$M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg$