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Katyanochek1 [597]
3 years ago
7

Which of the following is an example of a computer simulation?

Engineering
1 answer:
nata0808 [166]3 years ago
5 0

Answer:

b)prototype

Explanation:

_÷/#^'tixgixurykgdtusurdfjczut uxdkdktxkgxkt. it

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IM JI Suneou uo mm
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Answer: g

Explanation:

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

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3 years ago
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A steam turbine receives steam at 1.5MPa and 220oC, and exhausts at 50kPa, 0.75 dry. Neglecting heat losses and changes in kinet
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2 years ago
Are front-end engineers starting to decline in China?
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Yes. They are declining in China. Very fast

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3 years ago
Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate
uysha [10]

Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

k = Ae^{-E/RT} ------------------- euqation (1)

K= rate constant;

A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;

E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)

 or

k = 4.21 * 10⁻³((L/mol)s⁻¹)

or

k = 4.21 * 10⁻³(L/(mol.s))

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3 years ago
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