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ehidna [41]
3 years ago
6

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

ilograms. What is the kinetic energy of the system after the collision if the collision is elastic?
Physics
1 answer:
rewona [7]3 years ago
3 0

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

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A student of mass 51 kg wants to walkbeyond the edge of a cliff on a heavy beam ofmass 220 kg and length 11 m. The beam isnot at
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Answer:

4.465 m

Explanation:

Taken the beam to be uniform, then the center of gravity will act at the mid-point of the beam. The mid point = 11 / 2 = 5.5 m

since the boy can walk to the end of the beam without falling,

then the torque ( moment of the force) by beam = torque by students at the end of the beam

let the distance of the boy = x

220 ( 5.5 - x ) = 51 × x

1210 - 220 x = 51 x

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x = 1210 / 271 = 4.465 m

3 0
3 years ago
What step precedes the onset of hydrogen fusion in a star?
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6 0
3 years ago
An electron (restricted to one dimension) is trapped between two rigid walls 1.40 nm apart. The electron's energy is approximate
Svet_ta [14]

The quantum number of the energy state occupied by the electron is 10.

The given parameters;

  • <em>distance between the two rigid walls, L = 1.4 nm</em>
  • <em>energy of the electron, E = 19 eV</em>

The quantum number of the electron is calculated as follows;

E_n = \frac{h^2n^2}{8mL^2} \\\\

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
  • <em>m is mass of electron = 9.11 x 10⁻³¹ kg</em>
  • <em>E is the energy of the electron = 19 x 1.602 x 10⁻¹⁹ J</em>
  • <em>n is the quantum number of the energy state occupied by the electron.</em>

n^2 = \frac{8mL^2E_n  }{h^2} \\\\n = \sqrt{\frac{8(9.11\times 10^{-31} )(1.4 \times 10^{-9})^2 (19\times 1.602 \times 10^{-19})  }{(6.626\times 10^{-34})2} } \\\\n \approx 10

Thus, the quantum number of the energy state occupied by the electron is 10.

Learn more here:brainly.com/question/19426524

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