Respuesta :

dtscholar
The lack of info given to us requires that we use an energy approach. We start with electrical energy, U = -qV, where U is potential electrical energy, q is charge (which for a proton is e, which equals 1.6*10⁻¹⁹ C) and V is our potential difference. We can then use conservation of energy, assuming all our energy is converted to motion, so kinetic energy: K = 1/2m*v², where K is kinetic energy, m is the mass of the proton (1.67*10⁻²⁷ kg) and v is the velocity of the proton. We can set these energy terms equal to each other to get -qV = 1/2m*v², then we can solve for v, v = sqrt(-2qV/m), plug in the numbers, v = sqrt(-2(1.6*10⁻¹⁹)*(-1300)/(1.67*10⁻²⁷)) = 499,100 m/s, or about 1/10,000 the speed of light, which is well under the range of worrying about special relativity, so our method was correct.
johanrusli

The speed of a proton is about 5.0 × 10⁵ m/s

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Further explanation

Let's recall the Kinetic Energy formula:

[tex]\boxed {E_k = \frac{1}{2}mv^2 }[/tex]

Ek = kinetic energy ( J )

m = mass of object ( kg )

v = speed of object ( m/s )

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Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

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Given:

potential difference = ΔV = -1300 V

mass of proton = m = 1.67 × 10⁻²⁷ kilograms

charge of proton = q = 1.60 x 10⁻¹⁹ coulombs

initial speed of proton = v₁ = 0 m/s

Asked:

final speed of proton = v = ?

Solution:

[tex]Ep_1 + Ek_1 = Ep_2 + Ek_2[/tex]

[tex]qV_1 + \frac{1}{2}mv_1^2 = qV_2 + \frac{1}{2}mv_2^2[/tex]

[tex]q(V_1 - V_2 ) = \frac{1}{2}m( v_2^2 - v_1^2 )[/tex]

[tex]q( 0 - V ) = \frac{1}{2}m ( v^2 - 0^2 )[/tex]

[tex]-q \Delta V = \frac{1}{2} m v^2[/tex]

[tex]v^2 = -2mq \Delta V[/tex]

[tex]v = \sqrt { (-2q \Delta V) \div m }[/tex]

[tex]v = \sqrt { -2 \times 1.60 \times 10^{-19} \times (-1300) \div (1.67 \times 10^{-27})}[/tex]

[tex]v \approx 5.0 \times 10^5 \texttt{ m/s}[/tex]

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Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Energy

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