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Sliva [168]
1 year 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]1 year 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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Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

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We substitute

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2 years ago
From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:
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Answer:

The correct option is (B).

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(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU

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1 year ago
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Explanation:

            A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path  subjected to gravity force only.

           Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.

             In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.

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2 years ago
A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both
svetlana [45]

Answer:

7 deg

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k_{R} = spring constant for right spring = 33 Nm^{-1}

x_{L} = stretch in the left spring

x_{R} = stretch in the right spring

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\theta = Angle the rod makes with the horizontal

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Using equilibrium of force in vertical direction for right spring

k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m

Angle made with the horizontal is given as

\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg

3 0
2 years ago
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