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3 years ago

6

John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa

n company has agreed to give John a break on a deferred-payment plan that works as follows. John will not have to repay his loan for 5 years. During this “grace period” the loan obligation will compound at 4% per year. For the next 5 years, a monthly payment will be required, and the interest rate will be 0.5% per month. What will be John’s monthly payment over the 60-month repayment period?

1 answer:

IRISSAK [1]3 years ago

6 0

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= $35000\times (\frac{F}{P}, 4 \ percent,5 )$

= $35000\times 1.216 7$

= $42584.50$

Now,

time (t) will be:

= $5\times 12$

= $60 \ monthly \ payments$

So, monthly payment will be:

= $42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )$

= $42584.50\times 0.0193$

= $821.88$

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In a major human artery with an internal diameter of 5mm, the flow of blood, averaged over the cardiac cycle is 5cm3·s−1. The ar

antiseptic1488 [7]

9514 1404 393

Answer:

see attached

Explanation:

Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.

(5 cm³/s)(1.06 g/cm³) = 5.3 g/s

If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:

(5.3 g/s)/2 = 2.65 g/s

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The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...

(5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s

Downstream, we have half the volumetric flow and a smaller area.

(2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s

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3 years ago

Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25Â°C during this process a

AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS $_{air$ = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS $_{air$ = -40 kW / 298.15 K

ΔS $_{air$ = -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

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3 years ago

What is a robot’s work envelope?

Dmitry_Shevchenko [17]

Answer:

B

Explanation:

A robot's work envelope is its range of movement. It is the shape created when a manipulator reaches forward, backward, up and down. These distances are determined by the length of a robot's arm and the design of its axes. ... A robot can only perform within the confines of this work envelope.

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3 years ago

Define the following terms: data, database, DBMS, database system, database catalog, program-data independence, user view, DBA,

nalin [4]

Answer:

data - Any form of value in a column of a table in a relational database.

DBMS - Short for Database management system, which is a software that can be used to create, manipulate and view databases, e.g. MS SQL Server.

database system - Same as DBMS.

database catalog - Place where the metadata of a Database including its tables, users etc. exists, e.g. date created, size, number of columns etc. Also known as Data Dictionary.

program-data independence - Program-data independence refers to the design of keeping different levels of database (external, logical and physical) separate and loosely coupled from each other, for easier maintenance or modification work.

DBA - Short for Database Administrator, person responsible for maintaining the database. Its main responsibility is to keep the data clean and safe i.e. data doesn't contain wrong or invalid data, and is safe from viruses and is backed up.

end user - Anyone who is not directly interacting with a database, but through some software, like website or mobile application.

persistent object - An object of a class in a program, that interacts with the underlying database and is responsible for manipulating the database, the program is connected to.

transaction-processing system - A transaction process system is part of a software, responsible for making sure critical business transactions, like crediting or debiting money, either goes through cleanly or doesn't at all.

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3 years ago

A dryer is shaped like a long semi-cylindrical duct of diameter 1.5 m. The base of the dryer is occupied with water-soaked mater

Anestetic [448]

Answer:

0.0371 kg/s.m

Explanation:

From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)

Assuming the base surface of both ends of the cylinder is denoted by:

$A_1 \ and \ A_2$

Thus, using the summation rule, the view factor $F_{11$ and $F_{12$ is as follows:

$F_{11}+F_{12}=1$

Let assume the surface (1) is flat, the $F_{11} = 0$

Now:

$0+F_{12}=1$

$F_{12}=1$

However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder $A_2$ to the flat base surface $A_1$ ; we have:

$A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}$

Suppose, we replace DL for $A_1$ and

$A_2$ = $\dfrac{\pi D}{2}$

Then:

$F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\ =\dfrac{2}{\pi} \\ \\ =0.64$

Now, we need to employ the use of energy balance formula to the dryer.

i.e.

$Q_{21} = Q_{evaporation}$

But, before that; let's find the radian heat exchange occurring among the dome and the flat base surface:

$Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)$

where;

$\sigma = Stefan \ Boltzmann's \ constant$

$T_1 = base \ temperature$

$T_2 = temperature \ of \ the \ dome$

∴

$Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m$

Recall the energy balance formula;

$Q_{21} = Q_{evaporation}$

where;

$Q_{evaporation} = mh_{fg}$

here;

$h_{fg}$ = enthalpy of vaporization

m = the water mass flow rate

∴

$83899.15 = m \times 2257 \times 10^3 \\ \\ m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}$

6 0

3 years ago