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Tpy6a [65]
3 years ago
8

Sam promises to pay Sandy $2,000 in four years and another $3,000 four years later for a loan of $2,000 from Sandy today. What i

s the interest rate that Sandy is getting? Assume interest is compounded monthly. A. 14.75% B. 16.72% C. 15.10% D. 18.08%
Engineering
1 answer:
frozen [14]3 years ago
8 0

Answer:

the interest rate that Sandy is getting is (C.) 15.10%

Explanation:

Given data

in 4 year cash pay (p1)  = $2000

in 8 year cash pay (p2)  = $3000

time period (t1) = 4 years

time period (t2) =  8 years

loan value = $2000

To find out

interest rate

solution

first we know amount will be paid in first 4 year is

$2000 (1+r/100)^{12t}

$2000 (1+r/100)^{48}                  ...................1

now we calculate the next payment will paid after 4 year i.e.

$2000 (1+r/100)^{12t} - $2000  

$2000 (1+r/100)^{48}   - $2000  ..................2

after full time period of payment total amount will be paid by equation 1 and 2  i.e.

$3000 = $2000 (1+r/100)^{48} ×$2000 (1+r/100)^{48} - $2000

$3000 = $2000 ( (1+r/100)^{48} ×  (1+r/100)^{48} - 1 )   .....3

now we have solve   (1+r/100)^{48} this eqution

so we consider  (1+r/100)^{48} = A

so new equation will be by equation 3

$3000/$2000 = ( A - 1 ) × A

3/2 = A² - A

solve this equation we get  2A² - 2 A - 3 = 0 so A = 1.823

now we compute A again in  (1+r/100)^{48} = A

(1+r/100)^{48} = 1.823

so rate (r) = 1.258 % / month

and rate yearly = 1.258 ×12

the interest rate that Sandy is getting yearly 15.10 %

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Digiron [165]

Answer:

The final temperature of water is 381.39  °C.

Explanation:

Given that

Mass of water = 5 kg

Heat transfer at constant pressure Q = 2960 KJ

Initial temperature = 240 °C

We know that heat transfer at constant pressure given as follows

Q=mC_p\Delta T

We know that for water

C_p=4.187\ \frac{KJ}{kg.K}

Lets take final temperature of water is T

So

Q=mC_p\Delta T

2960=5\times 4.187(T-240)

T=381.39  °C

So the final temperature of water is 381.39  °C.

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3 years ago
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mafiozo [28]

Answer:

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What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he
inysia [295]

Given:

Temperature of water, T_{1} = -6^{\circ}C =273 +(-6) =267 K

Temperature surrounding refrigerator, T_{2} = 21^{\circ}C =273 + 21 =294 K

Specific heat given for water, C_{w} = 4.19 KJ/kg/K

Specific heat given for ice, C_{ice} = 2.1 KJ/kg/K

Latent heat of fusion,  L_{fusion} = 335KJ/kg

Solution:

Coefficient of Performance (COP) for refrigerator is given by:

Max COP_{refrigerator} = \frac{T_{2}}{T_{2} - T_{1}}

= \frac{267}{294 - 267} = 9.89

Coefficient of Performance (COP) for heat pump is given by:

Max COP_{heat pump} = \frac{T_{1}}{T_{2} - T_{1}}\frac{294}{294 - 267} = 10.89

6 0
3 years ago
A thick oak wall initially at 25°C is suddenly exposed to gases for which T =800°C and h =20 W/m2.K. Answer the following questi
Schach [20]

Answer:

a) What is the surface temperature, in °C, after 400 s?

   T (0,400 sec) = 800°C

b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s

c) What is the temperature, in °C, 1 mm from the surface after 400 s?

   T (1 mm, 400 sec) = 798.35°C

Explanation:

oak initial Temperature = 25°C = 298 K

oak exposed to gas of temp = 800°C = 1073 K

h = 20 W/m².K

From the book, Oak properties are e=545kg/m³   k=0.19w/m.k   Cp=2385J/kg.k

Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.

From energy balance: \frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t

Initial temperature wall = T_i

Surface temperature = T

Gas exposed temperature = T_{\infty}

6 0
3 years ago
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