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laila [671]
3 years ago
15

A -5.40nC point charge is on the x axis at x = 1.25m . A second point charge Q is on the x axis at -0.625m.

Physics
1 answer:
qwelly [4]3 years ago
5 0

Answe

a)  Q = 0.820 10⁻⁹ C ,   b)  Q = -3.52 10⁻⁹ C

Explanation:

The electric field is given by the formula

         E = k q / r²

where E is a vector quantity, so it must be added as a vector

          E_total = E₁ + E₂

let's look for the two electric fields

           E₁ = k q₁ / r₁²

           E₁ = 9 10⁹  5.4 10⁻⁹ / 1.25²

           E₁ = 31.10 N / C

           E2 = k Q / r₂²

           E2 = 9 10⁹ Q / 0.625²

           E2 = 23.04 10⁹ Q N / C           (1)

now we can solve the two cases presented

a) The total field is

            E_total = 50.0 N / C towards + x

since the test charge is positive the electric field E1 points to the right in the direction of the + x axis, so the equation is

            E_total = E1 + E₂

             E₂ = E_toal - E₁

             E₂ = 50.0 -31.10

             E2 = 18.9 N /C

With the value of the electric field we can calculate the charge (Q) using equation 1

             E₂ = 23.04 10⁹ Q

              Q = E₂ / 23.04 10⁹

              Q = 18.9 / 23.04 10⁹

              Q = 0.820 10⁻⁹ C

the charge on Q is positive

b) E_total = -50.0 N / C

              E_total = E₁ + E₂

              E₂ = E_total - E₁

              E2 = -50.0 - 31.10

               E2 = -81.10 N /C

we calculate the charge

             Q = E2 / 23.04 10⁹

             Q = -81.1 / 23.04 10⁹

              Q = -3.52 10⁻⁹ C

for this case the charge is negative

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The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

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Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

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2 years ago
If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radia
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Answer:

a) 5.5×10^17 Hz

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T = 2.60 \times 10^8 years

T = 2.60 \times 10^8 (365 \times 24 \times 3600) s

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Now the radius of the orbit of sun is given as

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3 years ago
Define the term overload.
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Have a good day and stay safe!

4 0
3 years ago
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