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mr Goodwill [35]
4 years ago
9

Explain how Newton's cradle demonstrates the law of conservation of energy.

Physics
1 answer:
AleksAgata [21]4 years ago
5 0

<u>Newton's cradle demonstrates the law of conservation of energy as follows:</u>

A tool named after the most popular scientist Sir Issac Newton, which utilizes a bunch of spinning balls to illustrate momentum and energy conservation, which is understood as Newton's cradle. Here the experiment involve hitting  of the stationary ball as one ball is raised and released at the end, transmitting a force through the stationary ball that forces the last sphere upward.

This illustrates the theory of momentum conservation (mass time speed), which says that the total momentum of the objects prior to the actual collision is equal to the total momentum of the entities after the collision, if two entities clash.

The constant banging of balls is also evidence of Newton's energy conservation law, which notes that energy can not be produced or destroyed but can switch ways. Also reveals this last aspect of the law very well, as it transforms one ball's potential energy into kinetic energy which is converted down the line of balls and eventually contributes to the last ball's upward swing.

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Riding his bike, dewayne can start from rest and get going at 12m/s in 4 seconds. Beth can get going 16m/s in 5 seconds. Who has
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Answer: Beth

Explanation: Dewayne (a) = 12/4 = 3m/s^2.

Beth (a) = 16/5 = 3.2 m/s^2.

So, Beth is the answer.

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3 years ago
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A ball is thrown at an angle of 45° to the ground. if the ball lands 87 m away, what was the initial speed of the ball? (round
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Consider the elementary gas-phase reversible reaction A 3C Pure A enters at a temperature of 400 K and a pressure of 10 atm. At
xxMikexx [17]

Answer:

  • 39%

Explanation:

The equilibrium equation is:

          A\rightleftharpoons 3C

The initial concentration of A can be calculated from the ideal gas equation:

                 pV=nRT\\\\n/v=p/(RT)\\\\C_A=\dfrac{10atm}{0.08206(atm-dm^3/K-mol)/times 400K}\\\\C_A=0.304mol/liter

Determine the conversion of substance A to substance C using an ICE table and the Kc constant:

             A               ⇄          3C

I            0.304                        0

C             - x                          + 3x

E          0.304 - x                   3x

         K_c=0.25=\dfrac{(3x)^3}{(0.304-x)}

Solve for x:

You need to use a graphing calculator:

  • 108x³ = 0.304 - x
  • 108x³ + x - 0.304 = 0
  • x ≈ 0.1195 mol/liter

Then:

           C_A=0.304mol/liter-0.1195mol/liter=0.1845mol/liter\\\\C_C=3\times 0.304mol/liter=0.912mol/liter

The equilbrium conversion is:

           \% = [0.304mol/liter-0.1845mol/liter]/(0.304mol/liter)\times 100

           \% \approx 39\%

3 0
3 years ago
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