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aleksklad [387]
3 years ago
7

Find the average net force.

Physics
1 answer:
Lerok [7]3 years ago
6 0

Answer:

<h2><em><u>given</u></em></h2>

<em>mass</em><em>=</em><em> </em><em>3</em><em>.</em><em>5</em><em> </em><em>kg</em>

<em>distan</em><em>ce</em><em> covered</em><em>=</em><em> </em><em>0</em><em>.</em><em>4</em><em>m</em>

<em>inital </em><em>speed</em><em> </em><em>=</em><em> </em><em>1</em><em>.</em><em>5</em><em> </em><em>m</em><em>/</em><em>sec</em>

<em>final</em><em> </em><em>speed</em><em> </em><em>=</em><em> </em><em>0</em><em> </em><em>m</em><em>/</em><em>sec</em>

<h2><u><em>To</em><em> </em><em>find</em></u><em> </em></h2>

<em>net</em><em> </em><em>force</em><em> </em>

<h2><em><u>solution</u></em></h2>

<em>formu</em><em>la</em><em> </em><em>for</em><em> </em><em>force</em><em> </em><em>:</em>

<em>\fbox{f = m.a}</em>

<u><em>In</em><em> </em><em>order</em><em> </em><em>to</em><em> </em><em>find</em><em> </em><em>the</em><em> </em><em>force </em><em>we</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>find</em><em> </em><em>the </em><em>acceleration</em><em>.</em></u>

<em><u>According</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>third</u></em><em><u> </u></em><em><u>equat</u></em><em><u>ion</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>Kin</u></em><em><u>ematics-</u></em>

<em><u>\bold{ {v}^{2} =  {u}^{2}  + 2as}</u></em>

<em>where</em><em>,</em>

<em>v</em><em>=</em><em> </em><em>fina</em><em>l</em><em> velocity</em>

<em>u</em><em>=</em><em>inita</em><em>l</em><em> Velocity</em>

<em>a</em><em>=</em><em>Acceleration</em>

<em>s</em><em>=</em><em>Distance</em><em> covered</em>

<em><u>put</u></em><em><u> </u></em><em><u>the </u></em><em><u>value</u></em><em><u> in</u></em><em><u> the</u></em><em><u> </u></em><em><u>abo</u></em><em><u>ve</u></em><em><u> </u></em><em><u>equa</u></em><em><u>tion</u></em>

<em>0</em><em>=</em><em> </em><em>(</em><em>1</em><em>.</em><em>5</em><em>)</em><em>²</em><em>+</em><em> </em><em>2</em><em>a</em><em> </em><em>x</em><em> </em><em>0</em><em>.</em><em>4</em>

<em>-</em><em>2.25</em><em>=</em><em> </em><em>0</em><em>.</em><em>8</em><em>a</em>

<em>a</em><em> </em><em>=</em><em> </em><em>-</em><em>2</em><em>.</em><em>2</em><em>5</em><em>/</em><em>0</em><em>.</em><em>8</em>

<em>a</em><em>=</em><em> </em><em>-</em><em> </em><em>2.8125</em><em> </em><em>m</em><em>/</em><em>sec²</em>

<em>now</em><em> </em><em>put</em><em> </em><em>the </em><em>value</em><em> </em><em>0</em><em>f</em><em> </em><em>accleration</em><em> </em><em>in</em><em> </em><em>formu</em><em>la</em><em> </em><em>of</em><em> </em><em>force</em><em> </em>

<h3><u><em>F</em><em> </em><em>=</em><em> </em><em>m</em><em>.</em><em>a</em></u></h3>

<em>F</em><em> </em><em>=</em><em> </em><em>3</em><em>.</em><em>5</em><em> </em><em>x</em><em> </em><em>-</em><em> </em><em>2</em><em>.</em><em>8</em><em>1</em><em>2</em><em>5</em>

<em>F</em><em>=</em><em> </em><em>-</em><em>9.84375</em><em> </em><em>N</em>

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There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center)
Aneli [31]

Answer:

q₁ =± 1.30 10⁻⁶ C  and   q₂ = ± 1.28 10⁻⁶ C

Explanation:

We will solve this problem with Coulomb's law

    F = K q₁q₂ / r²

Where the Coulomb constant is value 8.99 10⁹ N m² / C²

Let's apply this equation to our problem

Case 1

    F1 = k q₁ q₂ / r₁²

Where r₁ = 0.440 m and F1 = 0.0765 N

Case 2

The charges are the same

    F2 = k q q / r₂²

With r₂ = r₁ = 0.440 m, the spheres are fixed and the force is F2 = 0.100 N

When the spheres are joined with the wire, the charge is distributed, distributed and matched in the two spheres

    q₁ + q₂ = 2 q

Let's replace

    F2 = k ½ (q₁ + q₂) / r²

Let's write the two equations and solve the system of equations

    F1 = k q₁ q₂ / r²

    F2 = ½ k (q₁ + q₂) / r²    

    F1 r² / k = (q₁ q₂)

    F2 r² / k = (q₁ + q₂)/2

    q₁ = 2F2 r² / k - q₂

We substitute in the other equation

    F1 r² / k = (2F2 r² / k - q₂) q₂

    0 = -F1 r² / k + (2F2 r² / k) q₂ - q₂²

Let's solve the second degree equation

    F1 r² / K = 0.0765 0.440² / 8.99 10⁹

    F1 r² / K = 1.65 10⁻¹²

   (2F2 r² / k) =2  0.10 0.44² / 8.99 10⁹

    (2F2 r2 / k) = 4.30 10⁻¹²

    q₂² - 4.30 10⁻¹² q₂ + 1.65 10⁻¹² = 0

    q₂ = ½ {4.30 10⁻¹² ± √ [(4.30 10⁻¹²)² - 4 1.65 10⁻¹²]}

    q₂ = ½ {4.30 10⁻¹² ± 2,569 10⁻⁶}

    q₂ = ± 1.2845 10⁻⁶ C

Now we calculate q1

    F1 = k q₁ q₂ / r²

    q₁ = F1 r² / (k q₂)

    q₁ = 0.0765 0.440² / (8.99 10⁹ 1.2845 10⁻⁶)

    q₁ = 1.30 10⁻⁶ C

3 0
3 years ago
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