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LUCKY_DIMON [66]
3 years ago
11

An object at rest does not _____ and an object in motion does not _____, unless an _____ force acts upon it

Physics
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

An object at rest does not move and an object in motion does not change its velocity, unless an external force acts upon it

Explanation:

This statement is also known as Newton's first law, or law of  inertia.

It states that the state of motion of an object can be changed only if there is an external force (different from zero) acting on it: therefore

- If an object is at rest, it will remain at rest if there is no force acting on it

- If an object is moving, it will continue moving at constant velocity if there is no force acting on it

This phenomenon can be also understood by looking at Newton's second law:

F = ma

where

F is the net force on an object

m is the mass

a is the acceleration

If the net force is zero, F = 0, the acceleration of the object is also zero, a = 0: therefore, the velocity of the object does not change, and it will continue moving at the same velocity (which can be zero, if the object was at rest).

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Topic Gravitational force amd firld strength.. help me please
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The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

x \approx 6.74\,\mathrm m

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The best answer is D. The SI system, which is based on the metric system, has been more widely adopted than the English system, which is harder to use. In order to standardize measurements, the International System of Units or the SI units are commonly used around the world since they are more easy to understand and compare than the English unit.
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Read 2 more answers
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