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Natasha2012 [34]
3 years ago
8

An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts

on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 29 mm horizontally.

Physics
2 answers:
Mazyrski [523]3 years ago
6 0

Answer: 7.86×10^-4 mm

Explanation: find the attached file for the solution

Alex Ar [27]3 years ago
3 0

Answer:

 y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

        F = m a

        a = F / m

        a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

        a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

         x= 29 mm = 29 10⁻³ m

On the x axis

            v = x / t

            t = x / v

            t = 29 10⁻³ / 1.7 10⁷

            t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

            y = v_{oy} t + ½ a t²

            y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

            y = 77.74 10⁻⁵ m

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{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}

\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

<h3>☯ <u>By using formula of Lens</u> </h3>

\\

\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}

\\

\dashrightarrow\:\: \sf{ v = 30 \ cm}

\\

<h3>☯ <u>Now, Finding the magnification </u></h3>

\\

\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}

\\

\dashrightarrow\:\: \sf{m = -2}

\\

<h3>☯ <u>Hence</u>,\\</h3>

\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}

\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}

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