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ludmilkaskok [199]
3 years ago
5

An ink-jet printer steers charged ink drops vertically. Each drop of ink has a mass of 10-11 kg, and a charge due to 500,000 ext

ra electrons. It goes through two electrodes that gives a vertical acceleration of 104 m/s2. The deflecting electric field is _____ MV/m.
Physics
1 answer:
Sphinxa [80]3 years ago
4 0

Answer:

  E = 1.25 MV / m

Explanation:

For this exercise let's use Newton's second law

          F = m a

where the force is electric

          F = q E

we substitute

          q E = m a

          E = m a / q

indicate there are 500,000 excess electrons

          q = 500000 e

          q = 500000 1.6 10⁻¹⁹

          q = 8 10⁻¹⁴ C

the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²

         

let's calculate

          E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴

          E = 0.125 10⁷ V / m = 1.25 10⁶ V / m

          E = 1.25 MV / m

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An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
Nimfa-mama [501]

Explanation:

It is given that,

Speed of the electron in horizontal region, v=1.9\times 10^7\ m/s

Vertical force, F_y=4.9\times 10^{-16}\ N

Vertical acceleration, a_y=\dfrac{F_y}{m}

a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}  

a_y=5.37\times 10^{14}\ m/s^2..........(1)

Let t is the time taken by the electron, such that,

t=\dfrac{x}{v_x}

t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}

t=1.26\times 10^{-9}\ s...........(2)

Let d_y is the vertical distance deflected during this time. It can be calculated using second equation of motion:

d_y=ut+\dfrac{1}{2}a_yt^2

u = 0

d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2

d_y=0.000426\ m

d_y=0.426\ mm

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Which process repeatedly moves matter between different forms on Earth?
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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

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