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2 years ago

5 examples of neutons 3rd law of motion

Physics

2 answers:

olga55 [171]2 years ago

6 0

1). When you push on the wall with your hand, the wall pushes on your hand with an equal force.

2). You stand on the ground, and the ground pushes up on you with equal force.

3). When you bat the ball, the ball exerts an equal force on the bat.

4). When the horse pulls the wagon with 500 pounds of force, the wagon pulls the horse with 500 pounds of force. The source of that force is the horse, of course.

5). When the Earth's gravity pulls on you, the force is called "your weight on Earth". At the same time, your gravity pulls on the Earth, and that force should be called "the Earth's weight on you". The forces are equal ! If you weigh 75 pounds on Earth, then the Earth weighs 75 pounds on you.

And by the way, Sir Isaac spelled his name like "Newton".

Lemur [1.5K]2 years ago

3 0

Examples of Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.

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Minchanka [31]

Heterogeneous mixture

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3 years ago

Read 2 more answers

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

5 0

3 years ago

Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be ______

kotykmax [81]

Answer: The focal length of the cornea-lens system in his eye must be LESS THAN the distance between the front and back of his eye.

Explanation:

The human eye the front part of the eye is the CORNEA. This is the tough white transparent part of the eye that helps in the refraction of light rays. While the backside of the eye is the RETINA. This is the part of the eye when images are focused.

When a normal eye is at rest, parallel rays from a distant object are focused on the retina. The ability of the eye - lens to focus points at different distances on the retina is known as accomodation. The adjustment of the eye lens to focus objects of varying distances is brought about by the ciliary muscles. The have the ability to change the shape of the eye which leads to change in focal length.

When a person with normal vision looks at a distant object at infinity, the lens brings parallel rays to focus on the retina. Thus, the furthest point which the eye can see distinctly is called the far point of the eye and it's infinity for a normal eye. But Joe was able to focus his eye on the tree, meaning that the tree was within his near point. This is the nearest point at which an object is clearly seen. Therefore, when the effective focal length of the cornea-lens system changes, it changes the location of the image of any object in one's field of view.

5 0

2 years ago

Please help willing to give brainliest

-Dominant- [34]

Answer:

sound water because sound is the most important electronic wave for water

6 0

2 years ago

How much would a 70.0-kg person weigh on Mercury?

Natalija [7]

As weight =w =mg

g= gravitational acceleration on mercury = 3.7m/sec2

Mass of person =m= 70 kg

So w =(70kg)(3.7m/sec2)

w= 259 kgm/sec2

W= 259 N

6 0

2 years ago

5 examples of neutons 3rd law of motion​

5 examples of neutons 3rd law of motion