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Tju [1.3M]
3 years ago
15

What does Newton's second law of motion describe?

Physics
2 answers:
wolverine [178]3 years ago
7 0

Answer:

As the other person said it, relationship between force, mass, and acceleration.

Explanation:

Lyrx [107]3 years ago
3 0
Newton's second law of motion describes the relationship
between force, mass, and acceleration.

These days, we usually write it as

                   Force  =  (mass) x (acceleration) .
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What is it that makes a magnet different from a piece of iron that is not mangetic?​
Alisiya [41]

Answer:

Under normal conditions, a magnetic material like iron doesn't behave like a magnet because the domains don't have a preferred direction of alignment. On the other hand, the domains of a magnet (or a magnetized iron) are all aligned in s specific direction.

7 0
2 years ago
For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 81 N acts on
tensa zangetsu [6.8K]

Answer:

18 N

Explanation:

You have to do:

81 N: 9 ms2= x: 2 ms2

6 0
3 years ago
What are the charges of the subatomic particles?
kotegsom [21]
A particle smaller than an atom or a <span>cluster of such particles </span>
3 0
3 years ago
3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of
Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

5 0
1 year ago
A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu
malfutka [58]

Answer:

Acceleration, a=-2.48\ m/s^2

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

v^2-u^2=2as

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}

a=-2.48\ m/s^2

So, the acceleration on the rough ice -2.48\ m/s^2 and negative sign shows deceleration.

8 0
3 years ago
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