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Setler79 [48]
1 year ago
3

Describe the mechanical energy and the thermal energy of a bouncing basketball

Physics
1 answer:
miskamm [114]1 year ago
8 0

Answer:

when the ball is bounced gravitationally energy is released (more info below)

Explanation:

When a force is applied down to the ball, the gravitational potential energy is converted into kinetic energy before it hits the earth. Cinetic energy is converted into vibration, thermal, and static electrical energy, before the momentum of the ball pushes the ball back up into the hands of the people.

hope this helped!

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The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the
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F = 5.93×10^{13}\:\text{N}

Explanation:

Given:

m_1= 2×10^{16}\:\text{kg}

m_2= 4×10^{22}\:\text{kg}

r = 30000\:\text{km} = 3×10^7\:\text{m}

Using Newton's universal law of gravitation, we can write

F = G\dfrac{m_1m_2}{r^2}

\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}

\:\:\:\:= 5.93×10^{13}\:\text{N}

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A container with (15.0 + A) g of water at (8.0 + C) oC is placed in a freezer. How much heat must be removed from the water to t
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Answer:

Q_{T} = - 9.37 KJ

Explanation:

In this exercise the box has a caloric energy that must be eliminated. In several stages:  

First you must go from the current temperature to zero centigrade (T = 0ºC) which is the freezing point of water, it is heat we will call it Q₁  

Second you must lose heat without changing the temperature, it is changing state, latent heat Q_{L}

Third, the ice temperature should be lowered from zero degrees to T = - (5 + B) = - (5 +6) = -11 ° C, we will call it Q₂  

The heat removed is the sum of these heats  

    Q_{T} = Q₁ + Q_{L} + Q₂  

Let's calculate each one. For this you have the two equations is calorimetry.  

     Q = m c_{e} (T_{f}-T₀)  

     Q = m L                               change of state  

Let's calculate Q1, the data that give data is the mass m = 15.0 + A = 15.0 + 6 = 21.0 g, the initial temperature T₀ = 8.0 + C = 8.0 + 8 = 16ºC until the final temperature T_{f} = 0ºC  

Let's be careful because the specific heat is mixed in units, we must place the correct units so that they are simplified; this is the mass in grams, the temperature in degrees Kelvin  

    Q₁ = m c_{e} (T_{f}-T₀)  

    (T_{f}-T₀) = (273.15-289.15) = (0-16) = -16K  

    Q₁ = 21 4,186 (-16)  

     Q₁ = -1406.5 J  

Let's calculate the latent heat  

    Q_{L} = ±m L  

    Q_{L} = -21 333  

   Q_{L} = -6993 J  

The negative sign was selected because heat is being lost  

Calculate Q2  

   T = -11º C  

   Q₂ = 21 4,186 (-11 -0)  

   Q₂ = -966.97 J  

The total heat is  

   Q_{T} = -1406.5 - 6993 - 966.97  

   Q_{T} = - 9366.5 J  

The negative sign indicates that the heat is giving up.  

Reduce to kJ  

Q_{T} = -9366.5 J (1kJ / 1000J)  

Q_{T} = - 9.37 KJ

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