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Butoxors [25]
3 years ago
15

A student walks to school at a speed of 1.2 m/s. if the students mass is 53kg, what is the students kinetic energy

Physics
1 answer:
Ludmilka [50]3 years ago
3 0
Kinetic energy = 0.5 x mass x velocity squared
= 0.5 x 53 x (1.2)^2
38.16J
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An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect
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Answer:

The distance is d =1.66*10^{-9}m

Explanation:

From the question we are told that

         The initial speed of the  electron is v_i  = 3.2 *10^5 m/s

         The mass of electron is m = 9.11*10^{-31}kg

         Let d be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

         Let KE_i be the initial kinetic energy of the electron \

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  The energy at the initial position of the electron = The energy at the final position of the electron

      i.e

             KE_i +U_1 = KE_d + U_2

U_1 \ and \ U_2 are the potential energy at the initial  position of the electron and at distance d of the electron to the proton

                Here U_1 = 0

So the equation becomes

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Here q_1 \ and  \ q_2 are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

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i.e q = 1.602 *10^{-19}C

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             \frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}

            \frac{1}{2} mv_i^2  = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}

           3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}

Making d the subject of the formula

           d = \frac{2k(q)^2}{3mv_i^2}

              = \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}

              =1.66*10^{-9}m

             

           

         

                 

   

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