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Nina [5.8K]
3 years ago
10

A motorcycle has a constant acceleration of 3.49 m/s2. Both the velocity and acceleration of the motorcycle point in the same di

rection. How much time is required for the motorcycle to change its speed from (a)29.0 to 39.0 m/s, and (b)59.0 to 69.0 m/s?
Physics
1 answer:
Vilka [71]3 years ago
7 0

Answer:

(a)2.865 s

(b)2.865 s

Explanation:

We are given that

Acceleration,a=3.49 m/s^2

a.Initial speed,u=29 m/s

Final speed,v=39 m/s

We know that

t=\frac{v-u}{a}

Using the formula

t=\frac{39-29}{3.49}=2.865 s

b.Initial speed,u=59 m/s

Final speed,v=69 m/s

Again using the formula

t=\frac{69-59}{3.49}=2.865 s

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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:

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         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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  Considering that energy is conserved,

  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

                   \frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2  + \frac{kq_1 q_2}{d}

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           v_d is the velocity at distance d from the proton = 2v_i

  So the equation becomes

             \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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A circuit contains a single 270-pf capacitor hooked across a battery. it is desired to store four times as much energy in a comb
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