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3 years ago

Describe the rise and fall of a basketball using the concepts of kinetic energy and potential energy

1 answer:

Colt1911 [192]3 years ago

8 0

The equilibrium of the basketball makes it bounce lower every time. If it had perfect equilibrium, then it would bounce up to the same height every time. The kinetic energy is from when the ball hits the ground and is bouncing back up. But gravity will soon make it potential when it stays on the ground and when it stops rolling. When it rolls, the kinetic energy keeps it rolling until it gets so weak, there is only potential energy left to keep it stat

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How does the temperature of water change when it is heated on a stove top and then begins to boil?

guapka [62]

Answer:

Explanation:

First the water heats up to the boiling point ( temp increases)

then, as it boils it remains at constant temp ( boiling point)

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2 years ago

Which of the following is true about a planet orbiting a star in uniform circular

Mnenie [13.5K]

The velocity vector of the planet points toward the center of the circle is the following is true about a planet orbiting a star in uniform circular motion.

A. The velocity vector of the planet points toward the center of the circle.

<u>Explanation:</u>

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the circle.

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3 years ago

I’m designing a kitchen for a person in a wheel chair, which accommodation is least important? A. Provide turn around space for

Valentin [98]

The least important accommodation would be C.

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3 years ago

Read 2 more answers

What are some of the forces that are found in skateboarding?

Dominik [7]

Normal force, friction force, gravitational force

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3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

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3 years ago