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Paul [167]
3 years ago
11

you see: elián you hear: es el ocho, cuarenta y tres, cero, ocho, treinta y cinco. you write: 843-0835

SAT
1 answer:
Vesna [10]3 years ago
4 0

Elian's number is 843-0835 (in numbers) and eight, forty-three, zero, eight, thirty-five (in English)

The translation of the numbers from English to Spanish is as follows:

  1. Number - Name in English - Name in Spanish
  • 0 - Zero - Cero
  • 1 - One - Uno
  • 2 - Two - Dos
  • 3 - Three - Tres
  • 4 - Four - Cuatro
  • 5 - Five - Cinco
  • 6- Six - Seis
  • 7 - Seven - Siete
  • 8 - Eight - Ocho
  • 9 - Nine - Nueve
  • 10 - Ten - Diez

In the case of the largest numbers, for example twenty, thirty and forty the translation is as follows:

  • 20 - Twenty - Veinte
  • 25 - Twenty-five - Veinticinco
  • 30 - Thirty - Treinta
  • 35 - Thirty-five - Treinta y cinco
  • 40 - Forty - Cuarenta
  • 48 - Forty-eight - Cuarenta y ocho

According to this explanation, the number we hear is:

  • ocho (8), cuarenta y tres (43), cero (0), ocho (8), treinta y cinco (35).

Then the number in English would be:

  • Eight, forty-three, zero, eight, thirty-five.

Note: This question is incomplete because the question is missing. The question is:

- Cuál es el número de Elián?

- El número de Elián es: ocho, cuarenta y tres, cero, ocho, treinta y cinco.

- Ya lo escribí, gracias

How do you write Elian's number in numbers and in English?

Learn more in: brainly.com/question/11611530

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The net electric field is the vector sum of the components of the electric

field produced by the two charges.

The values of the magnitude and direction of the net electric field at the origin (approximate values) are;

  • 131.6 N/C
  • 12.6 ° above the negative x–axis

<h3>How are the net electric field magnitude and direction calculated?</h3>

The possible questions based on a similar question posted online are;

(a) The net electric field at the origin.

The electric field due to charge q₁ is given as follows;

\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}

Which gives;

\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C

\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C

Which gives;

\vec{E}_1 = \mathbf{21.6 \, N/C  \cdot \hat x +  28.8 \, N/C \hat y}

\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C

Therefore;

\vec  {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)

\vec  {E} = \mathbf{\left( -128.4 \, N/C  \right) \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)}

The magnitude of the net electric field is therefore;

E = \sqrt{(-128.4^2 + 28.8^2)} ≈ 131.6

  • The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>

(b) The direction of the net electric field at the origin.

  • The \ direction \ is \ arctan \left(\dfrac{28.8}{-128.4} \right) \approx \underline{ 12.6^{\circ}} \ above \ the \ negative \ x-axis

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