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Sagot :
Answer:
79.20 joule
Explanation:
1.
A 0.10kg stone is thrown from the edge of an ocean cliff with an initial speed of 21m/s. When it strikes the water below, it is traveling at 45m/s. What is the change in kinetic energy of the stone?
ΔEk = m/2*(V2^2-V1^2) = 0.05*(45^2-21^2) = 79.20 joule
The change in the kinetic energy of the stone is 81.25 J.
The given parameters;
- mass of the stone, m = 0.1 kg
- initial speed of stone, u = 20 m/s
- final velocity of the stone, v = 45 m/s
The initial kinetic energy of the stone is calculated as follows;
K.E = ¹/₂mu²
K.E₁ = ¹/₂ x 0.1 x 20²
K.E₁ = 20 J
The final kinetic energy of the stone is calculated as follows;
K.E₂ = ¹/₂ x 0.1 x 45²
K.E₂ = 101.25 J
The change in the kinetic energy of the stone is calculated as follows;
ΔK.E = K.E₂ - K.E₁
ΔK.E = 101.25 J - 20 J
ΔK.E = 81.25 J
Thus, the change in the kinetic energy of the stone is 81.25 J.
Learn more here:https://brainly.com/question/15598542
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