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ivann1987 [24]
2 years ago
7

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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Use the diagram to explain how convection occurs inside the earth. What could convection cause to occur on earths surface
liubo4ka [24]

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2 years ago
Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 2
crimeas [40]

Answer:

α = 13.7 rad / s²

Explanation:

Let's use Newton's second law for rotational motion

         ∑ τ = I α

         

we will assume that the counterclockwise turns are positive

         F₁  0 + F₂ R₂ - F₃ R₃ = I α

give us the cylinder moment of inertia

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        α = (F₂ R₂ - F₃ R₃)  \frac{2}{M R_2^2}

let's calculate

        α = (24  0.22 - 13  0.10) \frac{2}{12 \ 0.22^2}2/12 0.22²

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6 0
2 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
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