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Natalka [10]
3 years ago
6

Find the vector v with the given magnitude and the same direction as u. Magnitude ||v|| = 10 Direction u = ⟨-3. 4⟩.

Physics
1 answer:
Alisiya [41]3 years ago
4 0

Answer:

(-6,8)

Explanation:

Since v has direction as u, it must be a multiple n of u.

\sqrt{(-3n)^+(4n)^2}=10\\9n^2+16N^2 =100\\n^2=4\\n=2

therefore v = (3×2, 4×2) = (-6,8)

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in positive numbers less than 1, the zeros between the decimal point and a non-zero number are _______ significant?
DedPeter [7]

Answer:

Explanation:

If a number of less than 1, then the number has a decimal point like

0.085, 0.008 e.t.c.

The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.

But if there a zero between the none zero e.g. 0.0087056

Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6

But if the zero is not in between the none zero digit, then the zero is insignificant

E.g 0.05800

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So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.

8 0
3 years ago
A +5.00 pC charge is located on a sheet of paper.
emmainna [20.7K]

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = q_{int} /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

       

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

   

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

 

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

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3 years ago
What types of bonds do you think the good conductors of electricity have?
mariarad [96]
This what i found hope it helps

6 0
3 years ago
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