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Tpy6a [65]
2 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]2 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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Answer:

a. 130.73 atm

b. 102.62 atm

c. 87.1 atm

Explanation:

See the attached pictures.

6 0
3 years ago
Are ocean currents always cold
swat32

Answer:

The surface ocean currents have a strong effect on Earth's climate. ... However, these areas do not constantly get warmer and warmer, because the ocean currents and winds transport the heat from the lower latitudes near the equator to higher latitudes near the poles.

8 0
3 years ago
7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute
Crank

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

<h3>How to draw Shear Force and Bending Moment Diagram?</h3>

A) We can see the beam loaded in the first image attached.

For the shear diagram, let us calculate the shear from point load to point load.

From A to C, summing vertical to zero gives; ∑fy = 0: -20 - V = 0

V = -20 KN

From C to D, summing vertical to zero gives; ∑fy = 0: -20 + 48 - V = 0

V = 28 KN

From D to E, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - V = 0

V = 8 KN

From E to B, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - 20 - V = 0

V = -12 KN

For the bending moment diagram, let us calculate the bending moment from point load to point load.

At point A, the bending moment would be zero. Thus, M_A = 0 KN.m

At point C, taking moment about point C and equating to zero gives;

M_C = 0. Thus; 20(0.225) + M = 0

M = -4.5 KN.m

At point D, taking moment about point D and equating to zero gives;

M_D = 0. Thus; 20(0.525) - 48(0.3) + M = 0

M = 3.9 KN.m

At point E, taking moment about point E and equating to zero gives;

M_D = 0. Thus; 20(0.75) - 48(0.525) + 20(0.225) + M = 0

M = 5.7 KN.m

At point B, taking moment about point E and equating to zero gives;

M_E = 0. Thus; 20(1.05) - 48(0.825) + 20(0.525) + (20 * 0.3) + M = 0

M = 2.1 KN.m

2) From the attached diagrams, we can deduce that;

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

Read more about shear force & bending moment diagram at; brainly.com/question/14834487

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4 0
1 year ago
Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph
kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.  

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

Here, density of sea water is\rho_{sw}, surface gravity is S.G and density of water is \rho_{w}.

Substitute all the values in the above equation as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

1.025=\frac{\rho_{sw}}{1000}

\rho_{sw}=1025 kg/m³.

Step2

Difference in pressure is calculated as follows:

\bigtriangleup p=rho_{sw}gh

\bigtriangleup p=1025\times9.81\times320

\bigtriangleup p=3217680 pa.

Or

\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})

\bigtriangleup p=3217.68 kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

6 0
3 years ago
Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

Since the losses are neglected thus applying this theorm between upper and lower porion we have

\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

6 0
3 years ago
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