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Ludmilka [50]
3 years ago
7

alguien me ayuda con una tarea? Está en mi perfil de matemática, porfavorrr regalaré corona pero porfavor​

Engineering
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

HUH?

Explanation:

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function summedValue = SummationWithLoop(userNum) % Summation of all values from 1 to userNum summedValue = 0; i = 1; % Write a
Alexeev081 [22]

Answer:

function summedValue = SummationWithLoop(userNum)

% Summation of all values from 1 to userNum

  summedValue = 0;

  i = 0;

  % use a while loop that assigns summedValue with the

  % sum of all values from 1 to userNum

  while(i <= userNum)

      summedValue = summedValue + i;

      i = i + 1;

  end

end

8 0
3 years ago
Calculate the number of atoms per cubic meter in Metal B (units atoms/m^3). Write your answer with 4 significant figures metal:
Iteru [2.4K]

Answer:

7,217*10^28 atoms/m^3

Explanation:

  • Metal: Vanadium
  • Density: 6.1 g/cm^3
  • Molecuar weight: 50,9 g/mol

The Avogadro's Number, 6,022*10^23, is the number of atoms in one mole of any substance. To calculate the number of atoms in one cubic meter of vanadium we write:

1m^3*(100^3 cm^3/1 m^3)*(6,1 g/1 cm^3)*(1 mol/50,9g)*(6,022*10^23 atoms/1 mol)=7,217*10^28 atoms

Therefore, for vanadium we have 7,217*10^28 atoms/m^3

6 0
3 years ago
A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro
Vikki [24]

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= \frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt

= \int\limits^a_b {(5-3t)} \, dt

5t - \frac{3t^2}{2} +c

v2 = 5x2 -  3x2 + c

= 10-6+c

= 4+c

s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c

S2 - S1

=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

5 0
2 years ago
A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu
bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

a.

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

b.

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0
4 years ago
python Write a program that asks a user to type in two strings and that prints •the characters that occur in both strings. •the
Yuliya22 [10]

Answer:

see explanation

Explanation:

#we first get the elements as inputs

x = input("enter string A :")

y = input("enter string B :")

#then we make independent sets with each

x = set(x)

y = set(y)

#then the intersection of the two sets

intersection = set.intersection(x,y)

#another set for the alphabet

#we use set.difference to get the elements present in x and not in y, and

#viceversa, finally we get the difference between the alphabet and the #intersection of the elements in our strings

z = set('abcdefghijklmnopqrstuvwxyz')

print('\nrepeated :\n',intersection)

print('differences :\n',' Items in A and not B\n',

set.difference(x,y),'\nItems in B and not A\n',

set.difference(y,x))

print('\nItems in neither :\n',set.difference(z,intersection))

8 0
3 years ago
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