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Liono4ka [1.6K]
2 years ago
10

A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener

gy store has been transferred into its kinetic energy store. If the ball is travelling at 20 m/s when it hits the ground, what height was it dropped from? (Assume that the gravitational field strength is 10 N/kg.)
The answer is 20 meters, how do I get that?
Physics
1 answer:
gogolik [260]2 years ago
4 0

U = mgh, Ek = 1/2*m*v^2

U = Ek (conservation of mechanical energy)

⇒ mgh = 1/2*m*v^2

∵g = 10, v = 20

⇒ 10h = 1/2*400

⇒ h = 20 (m)

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The expression for the center of mass is

                x_{cm} = 1 / M  ∑ x_{i} m_{i}

Where M is the total masses, mI and xi are the mass and position of each element of the system.

Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon

              x_{cm} = 1 / (m₁ + m₂)   (0+ m₂ x₂)

Let's reduce the magnitudes to the SI system

             m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg

             m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg

             d = 128 pm = 128 10⁻¹² m

The equation for the center of mass is

               x_{cm} = m₂/ (m₁ + m₂)   d

b) let's calculate the value

            x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷    128 10-12

            x_{cm} = 52.97 10⁻¹² m

            x_{cm} = 52.97 pm

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