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3 years ago

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RKI Instruments borrowed $4,800,000 from a private equity firm for expansion of its facility for manufacturing carbon monoxide m

onitors. The company repaid the loan after 1 year with a single payment of $5,184,000. What was the interest rate on the loan?

1 answer:

Tju [1.3M]3 years ago

6 0

Answer:

n

Explanation:

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Sơ đồ nguyên lý của một hệ thống lạnh

Kobotan [32]

Answer:

A = 5h (B + b); solve for B.

2

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Is an ideal way for a high school student to see what an engineer does on a typical day but does not provide a hands-on experien

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Look at some engineering colleges and set up or join a public zoom meeting. For example, some colleges have sign ups for zoom calls on set dates and you‘re able to ask questions.

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A jet aircraft is in level flight at an altitude of 30,000 ft with an airspeed of 500 ft/s. The aircraft has a gross weight of 1

Naily [24]

Answer:

a) attached below

b) 0.0337

c) 2730.206 Ib

d) 2320.338 ft/min

Explanation:

<u>a) Plot of the drag polar for this aircraft </u>

first we will calculate :

Wing area (s) = Wing span (b) * Average chord length(c)

= 53.3 * 6 = 319.8 ft^2

Aspect ratio = b^2 / s = 8.883

K = 1 / $\pi$ eAR = 1 /

Drag polar ( Cd ) = 0.02 + 0.044 C^2L

attached below is a plot of the drag polar

<u />

Attached below is the detailed solution of the remaining part of the question

6 0

3 years ago

"Carbon 14 (C-14), a radioactive isotope of carbon, has a half-life of 5730 ± 40 years. Measuring the amount of this isotope lef

igor_vitrenko [27]

Answer:

The age of the bones is approximately 14172 years.

Explanation:

The age of the bones can be determinated using the following decay equation:

$N_{(t)} = N_{0}e^{-\lambda t}$ (1)

<u>Where:</u>

N(t): is the quantity of C-14 at time t

No: is the initial quantity of C-14

λ: is the decay rate

t: is the time

First, we need to find λ:

$\lambda = \frac{ln(2)}{t_{1/2}}$

<u>Where:</u>

t(1/2): is the half-life of C-14 = 5730 y

$\lambda = \frac{ln(2)}{5730 y} = 1.21 \cdot 10^{-04} y^{-1}$

Now, we can calculate the age of the bones by solving equation (1) for t:

$t = \frac{-ln(\frac{N_{(t)}}{N_{0}})}{\lambda}$

We know that the bones have lost 82% of the C-14 they originally contained, so:

$N_{t} = (1 - 0.82)N_{0} = 0.18N_{0}$

$t = \frac{-ln(0.18)}{1.21 \cdot 10^{-04} y^{-1}}$

$t = 14172 y$

Therefore, the age of the bones is approximately 14172 years.

I hope it helps you!

3 0

3 years ago

A state of stress that occurs at a point on the free surface of the of a solid body is = 50 MPa σ x , =10 MPa σ y , and = −15 MP

posledela

Answer:

A) 5 MPa , 55 MPa

B) maximum stress = 55 MPa, maximum shear stress = 25 MPa

Explanation:

using the given Data

free surface of a solid body

α $_{x}$ = 50 MPa, α $_{y}$ = 10 MPa , t $_{xy}$ = -15 MPa

attached below is the detailed solution to the question

7 0

4 years ago