Question

Question :

If an electron is subjected to a force of 10^-25 N in an X-ray machine, then find out the time taken by the electron to cover a distance of 0.2 m. Take mass of the electron 10^-30kg.​

Answer 1

Answer:

$\tt {\pink{Given}}\begin{cases} \sf{\green{Force= {10}^{ - 25} \:N }}\\ \sf{\blue{Mass=10^{ - 30} \: kg }}\\ \sf{\orange{Distance=0.2 \: m}}\\ \sf{\red{Time \:taken=?}}\end{cases}$

• Firstly we need to find Acceleration of electron. So, as we know to calculate acceleration we will use below given formula :]

$\bigstar \: \: \sf Acceleration = \dfrac{Force}{Mass} \: \: \bigstar$

• Now, Putting the values which is given in the question in the above given formula we get :]

$:\implies \sf Acceleration = \dfrac{10^{-25}}{10^{-30}}$

$: \implies \sf Acceleration = 10 \: ms^{ - 2}$

$\underline{ \textsf{ Hence, the Acceleration of electron is \textbf{10 ms ^{-2} }}}.$

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• The time taken by the electron (t) to cover distance (s) of 0.2 m can be given by second equation of motion :]

$\bigstar \:\:\sf s = ut + \dfrac{1}{2} \: a{t}^{2}\:\:\bigstar$

$\tt {\pink{We \:have}}\begin{cases} \sf{\green{Distance\: (s) = 0.2 \:m }}\\ \sf{\blue{Initial \: velocity\: (u) = 0}}\\ \sf{\orange{Acceleration\: (a)=10 \: ms^{-2}}}\\ \sf{\red{Time \:taken=?}}\end{cases}$

$\dashrightarrow \: \: \sf 0.2 = 0 + \dfrac{1}{2} \times {10}^{5} \times {t}^{2}$

$\dashrightarrow \: \: \sf 0.2 \times 2={10}^{5} \times {t}^{2}$

$\dashrightarrow \: \: \sf 0.4 = {10}^{5} \times {t}^{2}$

$\dashrightarrow \: \: \sf {t}^{2} = 0.4 \times {10}^{ - 5}$

$\dashrightarrow \: \: \sf {t}^{2} = 4 \times {10}^{ - 6}$

$\dashrightarrow \: \: \underline{ \boxed{\sf t = 2 \times {10}^{ - 3} \: s }}$

$\therefore\underline{ \textsf{ Time taken by the electron to cover a distance of 0.2 m is \textbf{10 ms ^{-3} seconds}}}.$

Answer 2

Answer:

The time taken by the electron to cover a distance of 0.2 m is 2×10^-3 sec.

Explanation:

Given :-

• An electron is subjected to a force of $\sf{10^{-25}}$ N in an X-ray machine.
• Mass of the electron is $\sf{10^{-30}}$ kg.

To find :-

• The time taken by the electron to cover a distance of 0.2 m.

Solution :-

• Force (F) of electron = $\sf{10^{-25}}$ N
• Mass (m) of the electron =$\sf{10^{-30}}$ kg

Now find the acceleration of the electron.

Formula used :

${\boxed{\sf{F=ma}}}$

• [ Put values ]

$\implies\sf{10^{-25}=10^{-30}\times\:a}$

$\implies\sf{10^{-30}\times\:a=10^{-25}}$

$\implies\sf{a=\dfrac{10^{-25}}{10^{-30}}}$

$\implies\sf{a=10^5}$

Then the acceleration of the electron is $\sf{10^{5}}$ m/s².

★ Formula used :

${\boxed{\sf{S=ut+\dfrac{1}{2}at^2}}}$

• S = 0.2 m
• u = 0 m/s
• a = $\sf{10^5}$ m/s²
• t = ?

[put values ]

$\implies\sf{0.2=0\times\:t+\dfrac{1}{2}\times\:10^5\times\:t^2}$

$\implies\sf{0.2=\dfrac{10^5\:t^2}{2}}$

$\implies\sf{10^5t^2=0.4}$

$\implies\sf{t^2=4\times\:10^{-6}}$

$\implies\sf{t=2\times\:10^{-3}}$

Therefore, the time taken by the electron to cover a distance of 0.2 m is 2×10^-3 sec.

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