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Neporo4naja [7]
3 years ago
8

Which describes a force acting on an object? a car moving in a straight line a ball that moves when kicked a billiard ball that

rolls at a constant speed
Physics
2 answers:
Amiraneli [1.4K]3 years ago
7 0
<span>Force is needed to change an objects state of motion and it is always a part of a mutual action that involves another force. The force can be push or pull.</span>
A ball that moves when kicked <span>describes a force acting on an object.
</span><span>The force is a vector quantity, because it that needs both magnitude and direction in order to be defined.</span>
AleksandrR [38]3 years ago
3 0

Answer:

  • a ball that moves when kicked
  • a ball that rolls at a constant speed

Explanation:

<h3>When a car moves in a straight line</h3>

For this statement we don’t know HOW the car is moving, it could be at constant speed, it could be changing its speed, we don’t know. So, this statement is not a very good description of how a force acts.

<h3>a ball that moves when kicked </h3>

The ball starts at rest, and, after being kicked, is moving. When a force acts in an object, the object is<u> accelerated</u>. This means, the force changes the velocity at which the object moves. That means that the force made by the kick changed its speed. This is the perfect description of how a force works.

<h3>a billiard ball that rolls at a constant speed</h3>

This one is a little tricky. One could say that this is just like the car, moving at constant speed, but that is not true, cause the car its modeled as an particle. The mechanics of rolling are more complicated. You see, if a ball spins, and there is no friction, the ball spins in its place. For the ball to start rolling, we need to consider the static friction that enables to rolls. So, there is a force acting on the object, the static friction force.

Of course, one could argue that there are friction forces acting on the car, and there is pressure, and so on, and so on, but in physics we model the systems as simple as necessary, and we use to imagine the models, as we say "In the vacuum" even if this is not remotely true.

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Explanation:

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\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

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\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

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m=9.11\times 10^{-31}\ \text{kg}

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