All categories

3 years ago

A company intends to market a new product and it estimates that there is a 20% chance that it will be first in the market

Engineering

1 answer:

Stells [14]3 years ago

4 0

The EMV - Ending Market Value is given as:
$2,400,000.

<h3>How is the EMV Arrived At?</h3>

The EMV is given as:

BMV x (i + r); Where

BMV is the Beginning Market Value; and

r is the interest rateor percentage given.

Hence the EMV = 2,000,000 x ( 1 + 20%)

= 2,000,000 x 1.2

= $ 2, 400,000.

It is to be noted that the BMV is the Beginning Market Value which is the value of an investment at the start of the business period.

Learn more about Market Value at:

brainly.com/question/1350233

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A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as

Ulleksa [173]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 = 101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ = 254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP = 9196.41/(0.00055501 - 0.000055501) = 18410899.5 kPa

4 0

3 years ago

If noise levels are high enough that you have to raise

Nana76 [90]

It would be ten because your ten feet away

4 0

3 years ago

A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value

Dominik [7]

Answer:

Cc= 12.7 lb.sec/ft

Explanation:

Given that

m = 22 lb

g= 32 ft/s²

$m = \dfrac{22}{32}=0.6875\ s^2/ft$

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

$K=\dfrac{m}{x}=\dfrac{22}{0.375}$

K= 58.66 lb/ft

The damper coefficient for critically damped system

$C_c=2\sqrt{mK}$

$C_c=2\sqrt{0.6875\times 58.66}$

Cc= 12.7 lb.sec/ft

5 0

3 years ago

The coolant heat storage system:

Monica [59]

Answer:

Explanation:

This is because many things, such as pcs, over heat

8 0

3 years ago

To 3 significant digits, what is the temperature of water in degrees C, if its pressure is 350 kPa and the quality is 0.01

liq [111]

Answer:

138.9 °C

Explanation:

The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C

3 0

3 years ago