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Sliva [168]
3 years ago
15

A rubber ball with a mass 0.20 kg is dropped vertically from a height of 1.5 m above the floor. The ball bounces off of the floo

r, and during the bounce 0.60 J of energy is dissipated. What is the maximum height of the ball after the bounce?
Physics
1 answer:
Digiron [165]3 years ago
8 0
Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m
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v = 1.08 m/s

Explanation:

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The decrease in PE is

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h = 0.80 m /2 = 0.40 m

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ΔPE = 0.1176 J

The moment of inertia of the assembly is

I = 1/12*m*L² + (m1 + m2)*(L/2)²

I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²

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KE = ½Iω²

0.1176 J = ½ * 0.032kg·m² * ω²

ω = 2.71 rad/s

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C=\frac{\epsilon_0 A}{d}

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\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

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2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

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u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

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4 years ago
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