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anastassius [24]
3 years ago
6

The following force diagram represents Newton’s Third Law of Motion:

Physics
1 answer:
Julli [10]3 years ago
4 0

Answer:

<u>FALSE.</u>

Explanation:

Newton's third law states that :

  • <em>Every action has equal and opposite reaction</em>
  • <em>That is , the magnitude is the same but the directions are opposite</em>
  • <em>The action reaction forces DONOT operate on the same body.</em>

For example ,

If a block is kept on the ground , the action force is the normal force acting on it due to the ground. <em>BUT , NOTE THAT : the reaction force isn't the gravitational force on the body ! It is the normal force acting on the ground due to the block !</em>

Thus,

we conclude that action and reaction forces donot act on the same body and therefore , this case has the <u>answer : FALSE </u>

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3 years ago
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Suppose that you are holding a cup of coffee in your hand identify all forces on the cup and reaction to each force
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Answer:

Newton's third law.

Explanation:

•your hand of holding UP/DOWN the cup has a reaction with force

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What is the only thing that can pull a beam of light towards itself ? ​
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3 years ago
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The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

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the other angle that gives the same result is

       θ‘= 90 - θ

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       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
a car whose mass is 1000kg is traveling at a constant speed of 10m/s. Neglecting any friction how much force will the engine hav
AURORKA [14]
This next statement is a big deal.  It should be up on a board, surrounded
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classroom.   Although we never see it in our daily lives, it's fundamental to
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<em>Without friction, it doesn't take <u>ANY</u> force to keep a moving object
moving.  </em>
<em>Force is only required to <u>change</u> the object's speed, or to
<u>change</u> the direction </em>
<em>in which it's moving.</em>

The answer to the question is:  On a level road, and neglecting any friction,
the engine doesn't have to supply ANY force to keep the car going at the
same speed.
7 0
3 years ago
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