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Serga [27]
3 years ago
8

What energy transformations from eating an apple to riding your bike?

Physics
2 answers:
elixir [45]3 years ago
8 0

Well I belive it has something to do with kinetic energy and I think it depends on if ur going up or down hill
Eduardwww [97]3 years ago
3 0
First you eat the apple
then your stomach "smash" ti
so the usefull part goes to your cell
they transform it into glucose
that you transform in cinetic energy by riding a bike
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Explain the following behaviour of molecules water rises up in a harrow tubes but mercury which is also a liquid falls in a narr
Mandarinka [93]

Answer: find the answer in the explanation

Explanation:

The capillarity of water molecules is different from the mercury molecules.

What is capillarity ?

This is the tendency of a liquid substance to rise in a capillary tube.

Molecules water rises up in a harrow tubes because of the force of adhesion between the water molecules and the tube molecules is greater than the force of cohesion between the water molecules. This helps water to wet the tube and rise. While mercury which is also a liquid falls in a narrow tubes to level below the outside surface because the force of cohesion between the mercury molecules is greater than the force of adhesion between the mercury molecules and the tube molecules. Mercury does not wet.

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3 years ago
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
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Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

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