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leonid [27]
3 years ago
6

According to the amortization table, Demarco and Tanya will pay a total of in interest over the life of their loan.

Engineering
2 answers:
Ymorist [56]3 years ago
8 0

Answer:

(Interest rate/number of payments)*$170000= interest for the first month.

Interest amounts for all the months of repayment plus $170000=Total loan cost

Explanation:

Interest is the amount you pay for taking a loan from a bank on top of the original amount borrowed.

Factors affecting how much interest is paid are; the principal amount, the loan terms, repayment schedule, the repayment amount and the rate of interest.

The interest paid=(rate of interest/number of payments to make)*principal amount borrowed.

You divide the interest with number of payments done in a year where monthly are divided by 12.Multiplying it by loan balance in the first month which is your principal amount gives the interest rate to pay for that month.

You new loan balance will be= Principal -(repayment-interest)

Do this for the period the loan should take.

Add all the interest amount to original borrowed amount to get total cost of the loan after the period of time.

algol133 years ago
6 0

Answer:

the answer for the first one is 104,854 and for the second one its 274,854

Explanation:

just did the assinment and got it right hope this helps  

You might be interested in
The nuclear reactions resulting from thermal neutron absorption in boron and cadmium are 10B5 + 1 n0 ï  7Li3 + 4He2 113Cd48 + 1
kirill115 [55]

Solution :

The nuclear reaction for boron is given as :

$^{10}\textrm{B}_5 + ^{1}\textrm{n}_0 \rightarrow ^{7}\textrm{Li}_3 + ^{4}\textrm{He}_2$

And the reaction for Cadmium is :

$^{113}\textrm{Cd}_48 + ^{1}\textrm{n}_0 \rightarrow ^{114}\textrm{Cd}_48 + \gamma [5 \ \textrm{MeV}]$

We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.

6 0
3 years ago
Answer?...................
torisob [31]

Answer:

The correct option is;

c. Leaving the chuck key in the drill chuck

Explanation:

A Common safety issues with a drill press leaving the chuck key in the drill chuck

It is required that, before turning the drill press power on, ensure that chuck key is removed from the chuck. A self ejecting chuck key reduces the likelihood of the chuck key being accidentally left in the chuck.

It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable

Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.

5 0
3 years ago
A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the
myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm  ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²  

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If  τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I)   ⇒  V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒  Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0
3 years ago
The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery
Lynna [10]

Answer:

capacity  = 0.555 mAh

capacity  = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C  

capacity  = 2 × ampere × second   ...............1

put here value and we get

and 1 Ah = 3600 C

capacity  = \frac{2}{3600}

capacity  = 0.555 mAh

and

when it is in parallel than capacity will be

capacity = Q1 +Q2   ...............2

capacity  = 1800 + 1800

capacity  = 3600 mAh

3 0
3 years ago
Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres
natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}

\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}

\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}

G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

h₃=2105KJ/Kg

h₄=957.14KJ/Kg

The net power developed:

P_{net}=P_{Tur}-P_{Comp}=G_{m}((h_{3}-h_{4})-(h_{2}-h_{1}))

The rate of heat:

Q=G_{m}(h_{3}-h_{2})

The thermal efficiency:

\mu_{ther}=\frac{P_{net}}{Q}

3 0
3 years ago
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