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Neporo4naja [7]
3 years ago
6

Let r be a bank's interest rate in percent per year (APR). An initial amount of money P, also called as principal, will mature t

o an amount of P(1+r/100)^n after n years have passed. Write a Python program that takes P, r, and n as inputs from the user and provides as output the matured value after n years. For example if the user provides P, r, and n as 1000 0.95 and 5 .Your program must apply the interest-rate formula and provide 1048.4111145526908 as the output.
Engineering
1 answer:
Liono4ka [1.6K]3 years ago
7 0

Answer:

See explanation below.

Explanation:

For this case the program needs to take the inputs as P,r and n and the output would be as A and printed on the system. The code is:

# Inputs

P = float(input("Enter the present value : "))  

r = float(input("Enter your APR : "))  

n = float(input("Enter the number of years : ") )

# Output

A = P*(1 +(r/100))**n

print("The future values is:", A)  

And the result obtained is:

Enter the present value : 1000

Enter your APR : 0.95

Enter the number of years : 5

The future values is: 1048.4111145526908

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Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0
2 years ago
Which of the following is NOT true concerning the color of minerals? A. Some minerals have a consistent color, but many have a r
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Answer:

Option D

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Color is one of the physical properties of minerals. Many minerals have a wide range of colors but there are some minerals with one consistent color and such minerals are referred as monochromatic minerals for example azurite. Normally, the streak color tends to be less variable than the color of the whole mineral and impurities or minor chemical components in a mineral react and often control the display color of resultant mineral. Option D is incorrect since mineral's color don't reflect wavelengths of light absorbed by such minerals.

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5 0
1 year ago
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

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Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
3 years ago
Jodie bought some shirts for 6$ each marge brought some shirts for 8$ each
Alex_Xolod [135]

Answer:

you need more details but if you have to find the difference, its $2.00

Explanation:

8-6=2

3 0
3 years ago
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