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alexgriva [62]
3 years ago
14

A particle has 37.5 J of kinetic energy and 12.5 J of gravitational potential energy at one point during its fall from a tree to

the ground. An instant before striking the ground, how much mechanical energy-rounded to the nearest Joule-does the particle have? Ignore air resistance.​
Physics
1 answer:
Varvara68 [4.7K]3 years ago
3 0

Answer: 50J

Explanation:

Mechanical energy follows the same principles of kinetic energy and potential energy, it is conserved. So Ei = Ef.

Mechanical energy is the sum of ALL energy's. There is no friction, so its just kinetic plus potential.

37.5 + 12.5 = 50J

Since the particle has not touched the ground, it has not transferred any energy to the ground yet, therefore the mechanical energy must still be 50J; mostly in kinetic energy with a very small amount of potential because of the low height relative to the ground.

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A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she
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I = 17.4 \times 10^{-3} kg m^2

Part b)

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Part a)

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I_1 = \frac{2}{5}m_1r_1^2

I_1 = \frac{2}{5}(1.6)((\frac{0.196}{2})^2)

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now the moment of inertia for thin shell

I_2 = \frac{2}{3} m_2r_2^2

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now total inertia of the ball is given as

I = I_1 + I_2

I = 17.4 \times 10^{-3} kg m^2

Part b)

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4 0
4 years ago
A 64.0 kg pole vaulter running at 10.2 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar i
Lubov Fominskaja [6]

Answer:

h = 5.05 m

Explanation:

given,

mass of pole vaulter, m = 64 Kg

speed of the vaulter,V = 10.2 m/s

horizontal component of velocity in air, v = 1 m/s

height of the jump,h = ?

using energy conservation

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\dfrac{1}{2}mv_i^2 + m g h_i= \dfrac{1}{2}mv_f^2+m g h_f

initial height of the vaulter is equal to zero.

\dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2+gh_f

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h =\dfrac{10^2-1^2}{2\times 9.8}

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height of the jump is equal to 5.05 m.

4 0
4 years ago
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