What happens to an object's gravitational potential energy as it falls from some height?

Physics

1 answer:

saveliy_v [14]3 years ago

3 0

Answer:

Decreases

Explanation:

As an object falls from some height, the gravitational potential energy of the object decreases.

Potential energy is the energy due to the position of a body.

The potential energy is mathematically expressed as:

Potential energy = mass x acceleration due to gravity x height

So, potential energy is directly proportional to mass and height.

The higher the mass and height, the more the potential energy.

You might be interested in

How is light energy different from both sound and heat energy?

Xelga [282]

Answer:

Light is energy that travels in waves and is produced by hot, energetic objects, while sound and heat energy are created with vibration and radiation, convection, or conduction.

Explanation:

Light bulbs are hot, energetic objects. You know the light bulb needs energy because you have to turn the light switch on to provide electricity for it. The electricity flows through either a thin metal wire or a gas. The wire or gas glows and gives off light when heated.

6 0

3 years ago

What energy changes take place in a glowing electric bulb

ehidna [41]

Answer:

electrical energy change into heat and light energy.

8 0

2 years ago

The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig

Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

the upper and the lower channels are at the same pressure (the atmospheric pressure).
the velocity of water in the upper channel is zero (v₁ = 0 m/s).
y₁ = 2.00 m (height of the upper channel)
y₂ = 0.00 m (height of the lower channel)
g = 9.81 m/s²
ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂

⇒ A = (350 m³/s)/(6.264 m/s)

⇒ A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4 ⇒ D = 2*√(A/π)

⇒ D = 2*√(55.873 m²/π)