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2 years ago

The force created when the court pushes LeBron James upwards is equal to which force?

Physics

2 answers:

nika2105 [10]2 years ago

8 0

Answer:

When LeBron James jumps, he is driving force into the court. How is this force created? His force is created by the energy stored inside his muscles.

Explanation:

i hope this helps you very much

max2010maxim [7]2 years ago

7 0

Answer:

When LeBron James jumps, he is driving force into the court. How is this force created? Suggested answer: This force is created by the energy stored inside his muscles.

Explanation:

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PLEASE HELP ME What does a moving charge experience when it is near a magnetic field? static friction gravity a stationary charg

attashe74 [19]

A moving charge experiences a force as it moves close to a magnetic field present in a region.

Explanation:

A charged particle when moves near a magnetic field is forced to move in a circular path. If there is a straight moving charged particle enters the region of the magnetic field, it's path is changed from straight path to the curved path.

The motion of the charged path inside or near the magnetic field will be a circular path and this circular path is due to the force experienced by the charged particle in the magnetic field.

The force experienced by the a charged particle as it moves near the magnetic field is termed as the Lorentz force. The expression for the Lorentz force experienced by the charged particle in the magnetic field is given by:

$\boxed{F_m=Q(\overrightarrow{v}\times\overrightarrow{B})}$

Here, $F_m$ is the magnetic force, $Q$ is the amount of charge, $\overrightarrow{v}$ is the velocity vector and $\overrightarrow{B}$ is the magnetic field intensity in the region.

Thus, A moving charge experiences a force as it moves close to a magnetic field present in a region.

Learn More:

1. Which one is true for the electromagnetic spectrum brainly.com/question/1619496

2. What is the current in resistor R2 brainly.com/question/3051098

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Electromagnetism

Keywords:

moving, charge, straight, circular, path, magnetic, field, amount, force, velocity, intensity, Lorentz, expression, stationary.

6 0

3 years ago

A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.

34kurt

To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that

$F = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal Constant

M = Mass of Earth (or Bigger star)

m = Mass of Object (or smallest star)

r = Radius

From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.

The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.

When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.

Therefore the correct answer is B.

6 0

3 years ago

Read 2 more answers

Two forces P and Q act on an object. Which diagram shows the resultant of these two forces?

denis-greek [22]

The Answer: diagram C

7 0

3 years ago

A friction like force that oppose motion of objects moving through the air is why we like to called

Art [367]

Air resistance ------- - - -- - - - :)

4 0

3 years ago

Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about t

satela [25.4K]

1) At the top, the ball has more potential energy

2) Halfway through the fall, potential energy and kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) Potential energy at the top: 784 J

5) Potential energy halfway through the fall: 392 J

6) Kinetic energy halfway through the fall: 392 J

7) KInetic energy before hitting the ground: 784 J

Explanation:

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object above the ground

The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

For the bowling ball in the problem, when it sits on top of the building it has no kinetic energy (because its speed is zero, v = 0), therefore it has more potential energy than kinetic energy.

The total mechanical energy of the ball, which is the sum of the potential and the kinetic energy, is constant during the fall:

$E=PE+KE=const.$

When the ball is at the top, all its energy is potential energy, since the kinetic energy is zero:

$E=PE=mgH$

where H is the initial height.

When the ball is halfway through the fall, the height is H/2, so:

$PE=mg\frac{H}{2}$

which means that the potential energy is now half of the total mechanical energy: but since the total energy must be constant, this means that the kinetic energy is now also half of the total energy. Therefore, potential energy and kinetic energy are equal.

When the ball is just before hitting the ground, the height of the ball is now zero

h = 0

This also means that the potential energy is zero

PE = 0

Therefore, all the energy of the ball is now kinetic energy:

$KE=E$

which means that the kinetic energy is maximum, and therefore it is larger than the potential energy: this is because the ball accelerates during the fall, and therefore its speed is maximum just before hitting the ground.

The potential energy of the ball is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object above the ground

When the ball sits at the top, we have

m = 2 kg

$g=9.8 m/s^2$

h = 40 m

Therefore, the potential energy is

$PE=(2)(9.8)(40)=784 J$

The potential energy of the ball is given by

$PE=mgh$

where

m = 2 kg is the mass

$g=9.8 m/s^2$ is the acceleration due to gravity

When the ball is halfway through the fall, the height of the ball is

h = 20 m

Therefore, its potential energy is

$PE=(2)(9.8)(20)=392 J$

which is half of the initial potential energy.

The kinetic energy of the ball is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the ball

v is its speed

When the ball is halfway through the fall, we have:

m = 2 kg (mass of the ball)

v = 19.8 m/s (speed of the ball)

Therefore, the kinetic energy is

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

Which is equal to the potential energy.

The kinetic energy of the ball just before hitting the ground is

$KE=\frac{1}{2}mv^2$

where in this case,

m = 2 kg is the mass

v = 28 m/s is the speed of the ball

Therefore, kinetic energy is

$KE=\frac{1}{2}(2)(28)^2=784 J$

And we see that the kinetic energy of the ball just before hitting the ground is equal to the potential energy of the ball when it sits at the top: therefore, all the mechanical energy has converted from potential energy into kinetic energy.

Learn more about kinetic and potential energy:

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brainly.com/question/1198647

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3 0

3 years ago