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3 years ago

A 120 g coconut falls 12 m. What is the kinetic energy of the coconut just before it hits the ground?

1 answer:

5 0

Kinetic Energy (Ek) =
$m \times {v}^{2} \div 2$

Torricelli's Equation
g earth = 10m/s^2

${v}^{2} = {vo}^{2} + 2g \times delta(s) \\ {v}^{2} = 0 + 2 \times 10 \times 12 \\ {v}^{2} = 240m/s$

120g = 0.12 kg

Ek = 0.12 x 240/2
Ek = 28.8/2
Ek = 14.4 J

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A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

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Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

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mass, m = 1kg

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suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

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Derivate with respect to t

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(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

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