Answer: The monthly payment before the buydown is $71.3
The monthly payment after the buydown is $68.9
Explanation: The payment is compounding so we use compound interest;
A= P[1+(r/n)^nt]
Where;
A= Compounded amount
P = principal
r= interest rate per payment
n= number of payment per period
t= number of period.
NOTE: from our questions, the period is yearly and the payment is monthly. Therefore;
number of payment per period (n) is 12
number of payment period (t) is 10
P=$8000, r= 0.667% or 0.333%
FIND MONTHLY PAYMENT BEFORE BUYDOWN:
Step 1: find the Compounded amount to pay.
A= $8000[1+(0.00667÷12)^(12×10)]=
$8551.64 this is the total amount he has to pay for a period of 10years
Step 2: How much does he has to pay monthly for a period of 10year;
Therefore his payment will be for 120 months
$8551.64÷120= $71.3 monthly
FIND MONTHLY PAYMENT AFTER BUYDOWN:
Step 1: find the compounded amount to pay.
A= 8000[1+(0.00333÷12)^(12×10)=
$8270.85 this is the total amount he has to pay for a period of 10years
Step2: How much does he has to pay monthly for a period of 10year;
Therefore his payment will be for 120 months;
$8270.85÷120= $68.9 monthly
• Use the reference on "Practical Considerations for Rectangular Reinforced Concrete Beams"
• Include references to ACI code – see slides from second class
• Include references to Tables from Appendix A
• Draw a sketch of the reinforced concrete beam showing all dimensions, number and size of rebars, including stirrups.
Answer:
Beam of 25" depth and 12" width is sufficient.
I've attached a detailed section of the beam.
Explanation:
We are given;
Beam Span; L = 20 ft
Dead load; DL = 0.50 k/ft
Live load; LL = 0.65 k/ft.
Beam width; b = 12 inches
From ACI code, ultimate load is given as;
W_u = 1.2DL + 1.6LL
Thus;
W_u = 1.2(0.5) + 1.6(0.65)
W_u = 1.64 k/ft
Now, ultimate moment is given by the formula;
M_u = (W_u × L²)/8
M_u = (1.64 × 20²)/8
M_u = 82 k-ft
Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.
Effective depth of a beam is given by the formula;
d_eff = d - clear cover - stirrup diameter - ½Main bar diameter
Now, let's adopt the following;
Clear cover = 1.5"
Stirrup diameter = 0.5"
Main bar diameter = 1"
Thus;
d_eff = 25" - 1.5" - 0.5" - ½(1")
d_eff = 22.5"
Now, let's find steel ratio(ρ) ;
ρ = Total A_s/(b × d_eff)
Now, A_s = ½ × area of main diameter bar
Thus, A_s = ½ × π × 1² = 0.785 in²
Let's use Nominal number of 3 bars as our main diameter bars.
Thus, total A_s = 3 × 0.785
Total A_s = 2.355 in²
Hence;
ρ = 2.355/(22.5 × 12)
ρ = 0.008722
Design moment Capacity is given;
M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12
Φ is 0.9
f’c = 4,000 psi = 4 kpsi
fy = 60,000 psi = 60 kpsi
M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12
M_n = 220.03 k-ft
Thus: M_n > M_u
Thus, the beam of 25" depth and 12" width is sufficient.
Three activities that I can do on a daily basis that involve both metric units (SI units) and customary units are: measuring the length of a door with a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that calls for one teaspoon (customary unit) of baking soda, which can also be converted to four grams (SI unit); and weighing myself on a weighing scale, which can be measured in pound and kilogram (metric unit).
Answer: Three examples of activities that I can perform on a daily basis that involves both metric units (SI units) and customary units include: measuring the length of a door using a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that requires one teaspoon (customary unit) of baking soda, which could also be converted into four grams (SI unit); weighing myself on a weighing scale, which can be measured by pounds (customary unit) or kilograms (metric unit).
Explanation:I big brain :) (Not Really I Just Wanted To Help) I hope this helped! ;)
Answer:
Explanation:
The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )
The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797
The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95
The annual worth of land application = $ 121,560.59
The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )
The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607
The annual worth of land Incineration = $ 211,283.85
The annual worth of contract = $ 190,000
The annual worth of contract = $ 190,000
The land application has the least cost , hence it is preferred .
Answer: c) 450 kPa
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
where,
= initial pressure of gas = 150 kPa
= final pressure of gas = ?
= initial volume of gas = v L
= final volume of gas =
Therefore, the new pressure of the gas will be 450 kPa.
Answer:
The flow boiling is also classified as either external and internal flow boiling depending on whether the fluid is forced to flow over a heated surface or inside a heated channel. The two-phase flow in a tube exhibits different flow boiling regimes, depending on the relative amounts of the liquid and the vapor phases.
Answer:
Given:
operating temperature of heat engine,
Solution:
For a reversible cycle, maximum efficiency, is given by:
Now, on re designing collector, maximum temperature, changes to
, so, the new maximum efficiency,
is given by: