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eimsori [14]
1 year ago
10

Suppose an acorn with a mass of 3.17 g falls off a tree. At a particular moment during the fall, the acorn has a kinetic energy

of 0.119 J and a potential energy of 0.347 J with respect to the ground. How does the acorn's potential and kinetic energy change as it falls
Physics
1 answer:
denis-greek [22]1 year ago
6 0

Potential and kinetic energy both decrease with the acorn's falling potential and kinetic energy.

The acorn's potential energy is at its peak when it reaches the top of the tree, yet its kinetic energy is zero (i.e., it is not accelerating).

The height of the ball reduces along with the potential energy as the acorn tumbles down the tree, but the kinetic energy rises (energy due to motion)

The height will be 0 and the kinetic and potential energy will be zero at the ground. This demonstrates that as an item falls, both potential and kinetic energy are lost.

Learn more about Energy here

brainly.com/question/13881533

#SPJ4

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A plane travelling at 63 m/s[S] down a runway begins accelerating uniformly at 2.8 m/s?[S]. How far does it travel
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Answer:

270 m

Explanation:

Given:

v₀ = 63 m/s

a = 2.8 m/s²

t = 4.0 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (63 m/s) (4.0 s) + ½ (2.8 m/s²) (4.0 s)²

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Rounded to two significant figures, the displacement is 270 meters.

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How does a rubber rod become negatively charged through friction?
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A 4.00-kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6.00-kg block tha
Anna007 [38]

Answer:

v = 2.82 m/s

Explanation:

For this exercise we can use the conservation of energy relations.

We place our reference system at the point where block 1 of m₁ = 4 kg

starting point. With the spring compressed

        Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂

final point. When block 1 has descended y = - 0.400 m

        Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y

as there is no friction, the energy is conserved

       Em₀ = Em_f

       ½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y

       ½ k x² - m₁ g y = ½ m₂ v²

 

       v² = \frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y

let's calculate

        v² = \frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)

        v² = 2.7 + 5.23

        v = √7.927

        v = 2,815 m / s

using of significant figures

        v = 2.82 m/s

5 0
3 years ago
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