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Marrrta [24]
3 years ago
12

A sum of $500,000 will be invested by a firm two years from now. If money is worth 12%, what will be the worth of this investmen

t 10 years from now?
Engineering
1 answer:
notka56 [123]3 years ago
8 0

Answer:

investment 10 years from now is $1,238,000 .

Explanation:

given data

sum = $500,000

rate = 12% =0.12

total time = 10 year

solution

as present value After 2 years from now is $500,000

so time period is now = 8 year  ( 10 - 2 )

so we apply future value formula that is

Future value  = present value × (1+r)^{t}   ............1

put here value we get

Future value  = $500,000 × (1+0.12)^{8}  

Future value  = $500,000 × 2.476

Future value  = $1,238,000  

so investment 10 years from now is $1,238,000 .

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4. Which 2D shape on the left would be used to make the 3D shape on the right? (1 pt.)
dexar [7]

it would be a bc its a sqare?well 3d is like you can say a cube 2d is like flat

Explanation:

7 0
2 years ago
Three communications channels in parallel have independent failure modes of 0.1 failure per hour. These components must share a
ozzi

Answer:

the MTTF of the transceiver is 50.17

Explanation:

Given the data in the question;

failure modes = 0.1 failure per hour

system reliability = 0.85

mission time =  5 hours

Now, we know that the reliability equation for this situation is;

R(t) = [ 1 - ( 1 - e^{-0.1t )³] e^{-t/MTTF

so we substitute

R(5) = [ 1 - ( 1 - e^{-0.1(5) )³] e^{-5/MTTF = 0.85

[ 1 - ( 1 - e^{-0.5 )³] e^{-5/MTTF = 0.85

[ 1 - ( 0.393469 )³] e^{-5/MTTF = 0.85

[ 1 - 0.06091 ] e^{-5/MTTF = 0.85

0.9391 e^{-5/MTTF = 0.85  

e^{-5/MTTF = 0.85 / 0.9391

e^{-5/MTTF = 0.90512

MTTF = 5 / -ln( 0.90512 )

MTTF = 50.17

Therefore, the MTTF of the transceiver is 50.17

7 0
3 years ago
When in regular operation, ash and other debris should be removed _______ from the combustion chamber of a wood-burning heater.
Klio2033 [76]

Answer:

    Daily from the combustion chamber of a wood-burning heater.

<h3>Explanation:</h3>
  • As a wood stove heats up, it radiates heat through the walls and top of the stove
  • This radiant heat warms the immediate area and can be carried into other parts of the home via the home's natural airflow.
  • Electric or convection-powered fans can help circulate this heat to warm a larger area.

To learn more about it, refer

to brainly.com/question/23275071

#SPJ4

8 0
2 years ago
A single square-thread screw has an input power of 3 kW at a speed of 1 rev/s. The screw has a diameter of 40 mm and a pitch of
Sonbull [250]

Answer:

Axial Resisting Load, F = 31.24kN

Efficiency = 16.67%

Explanation:

Given

Input Power = P,in = 3kW = 3,000W

Speed, S = 1rev/s

Pitch, p = 8mm

Thread frictional coefficient = μt = 0.18

Collar frictional coefficient = μc = 0.09

Friction radius of collar, Rc = 50mm

First, we calculate the torque while the load is being lifted in terms of 'F'.

This is calculated by

T = ½FDm[1 + πDmμt]/[πDm - μtp]

By substituton.

T = ½F(40-4)[1 + π(40-4)0.18]/[π(40-4) - 0.18 * 8]

T = 18F(1 + 6.48π)/(36π - 1.44)

T = 3.44F.Nmm

T = 3.44 * 10^-3F Nm

Then we calculate the torque due to friction from the collar

T = Fμc * Rc

T = F * 0.09 * 50

T = 4.5F. Nmm

T = 4.5 * 10^-3F Nm

Then, we calculate the axial resisting load 'F' by using the the following power input relation.

P,in = Tw

P,in = (T1 + T2) * 2πN

Substitute each value

3,000 = (3.44 + 4.5) * 10^-3 * F * 10^-3 * 2 * π * 2

F = 3000/((3.44 + 4.5) * 10^-3 * 10^-3 * 2 * π * 2

F = 31,247.69N

F = 31.24kN

Hence, the axial resisting load is

F = 31.24kN

Calculating Efficiency

Efficiency = Fp/2πP

Efficiency = 2Fp/P,in

Substitute each value

Efficiency = 2 * 31,247.69 * 8 * 10^-3/3000

Efficiency = 0.166654346666666

Efficiency = 16.67%

8 0
4 years ago
Read 2 more answers
Describe experimental factors that could be modified, and unalterable properties of materials used.
Sphinxa [80]

Answer:

a. mechanical properties

b. thermal properties

c. chemical properties

d. electical properties

e. magnetic properties

Explanation:

a. The mechanical properties of a material are those properties that involve a reaction to an applied load.The most common properties considered are strength, ductility, hardness, impact resistance, and fracture toughness, elasticity, malleability, youngs' modulus etc.

b. Thermal properties such as boiling point , coefficient of thermal expansion , critical temperature  , flammability  , heat of vaporization , melting point ,thermal conductivity , thermal expansion ,triple point , specific heat capacity

c. Chemical properties such as corrosion resistance , hygroscopy , pH , reactivity , specific internal surface area , surface energy , surface tension

d. electrical properties such as capacitance , dielectric constant , dielectric strength , electrical resistivity and conductivity , electric susceptibility , nernst coefficient (thermoelectric effect) , permittivity  etc.

e. magnetic properties such as diamagnetism,  hysteresis,  magnetostriction , magnetocaloric coefficient , magnetoresistance , permeability , piezomagnetism , pyromagnetic coefficient

3 0
3 years ago
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