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dedylja [7]
3 years ago
9

A bob of mass of 0.18 kilograms is released from a height of 45 meters above the ground level. What is the value of the kinetic

energy gained by the bob at the ground level
Physics
1 answer:
goblinko [34]3 years ago
3 0
Plug into the equation K=1/2mv^2 sooo.. (1/2)(.18)v(you don't have velocity so use the equation for potential energy because you haveheight)(U=mgh so (.18kg)(9.8m/s)(45)=U (set U equal to K)  
(.18)(9.8)(45m)=1/2(.18)v^2
solve for v
don't forget to take the square root of v
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let the length of the beam be "L"

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AC = CD = AD/2 = L/2

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BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

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using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

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Making V the subject of the equation,

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Given: Q = 4.5  µC = 4.5×10⁻⁶ C, C = 3.5  µF = 3.5×10⁻⁶ F

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