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pishuonlain [190]
3 years ago
9

a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an

angle of 45 and is 51.0 m long/ What will its kinetic energy be wehn it reaches the bottom
Physics
1 answer:
igor_vitrenko [27]3 years ago
8 0

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is m=230\ kg

It reaches the steepest hill with speed of u=6.2\ km/h\ or \ 1.72\ m/s

Hill to bottom is 51 m long with inclination of 45^{\circ}

Height of the hill is h=51\sin 45^{\circ}=36.06\ m

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ

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Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

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