Find the acceleration of a train whose speed increases from 7m/s to 17 m/s in 120s

1 answer:

7 0

Initial speed of the train = 7 m/s
Final speed of the train = 17 m/s
Change of speed of the train = (17 - 7) m/s
   = 10 m/s
Time taken for the change of the speed of the train = 120 s
Then
Acceleration of the train = Change of speed of the train/Time taken for the change of speed
   = 10/120 m/s^2
   = 1/12 m/s^2
   = 0.083 m/s^2
So the acceleration of the train is 1/12 meter per second square or 0.083 m/s^2. I hope this is the answer you were looking for.

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

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Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron, $u_2=0$

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$r_2=2.3 cm=\frac{2.3}{100}=0.023 m$

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We have to determine the energy of the incident electron.

Mass of electron, $m=9.1\times 10^{-31} kg$

Charge on an electron, $q=-1.6\times 10^{-19} C$

Velocity, $v=\frac{Bqr}{m}$

Using the formula

Speed of electron, $v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0$

Speed of second electron, $v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}$

$v_2=1.5\times 10^8 m/s$

Kinetic energy of incident electron= $\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2$

Kinetic energy of incident electron= $0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J$

Kinetic energy of incident electron= $\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV$

1KeV=1000eV

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Answer:

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