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Travka [436]
4 years ago
14

Which situation is contrary to Newton’s first law of motion? An object at rest begins to move when an external force is applied.

A moving object has a constant velocity under the influence of balanced forces. An object at rest stays at rest as long as unbalanced forces act on it. A moving object slows down when an external force is applied.
Physics
2 answers:
vaieri [72.5K]4 years ago
7 0

Newton's first law establishes the following:

An object will remain at rest or with uniform rectilinear motion unless an external force acts on it.

Therefore, we have that a case contrary to Newton's first law is:

Unbalanced forces acting on an object at rest, cause the object to remain at rest.

It is false, because if the forces are unbalanced, then the object will not remain at rest because external forces act on the object.

Answer:

A situation that is contrary to Newton's first law of motion is:

An object at rest stays at rest as long as unbalanced forces act on it.

Dmitrij [34]4 years ago
6 0

C. An object at rest stays at rest as long as unbalanced forces act on it.

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Suppose the cyclist travels for a total time of <em>t</em> hours.

For 20 min = 1/3 hr, the cyclist does not move.

Over the remaining (<em>t</em> - 1/3) hr, the cyclist is moving at a constant speed of 22.0 km/hr, so that the cyclist would travel a distance of

<em>x</em> = (22.0 km/hr) • ((<em>t</em> - 1/3) hr) ≈ (22.0 km/hr) <em>t</em> - 7.33 km

If the cyclist's average speed over the total time <em>t</em> was 17.5 km/hr, then by the definition of average speed,

17.5 km/hr = <em>x</em> / <em>t</em>

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17.5 km/hr = ((22.0 km/hr) <em>t</em> - 7.33 km) / <em>t</em>

Solve for <em>t</em> :

17.5 km/hr = 22.0 km/hr - (7.33 km) / <em>t</em>

(7.33 km) / <em>t</em> = 4.5 km/hr

<em>t</em> = (7.33 km) / (4.5 km/hr)

<em>t</em> ≈ 1.62963 hr

Then the distance the cyclist traveled must have been

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Alternatively, as soon as you arrive at

17.5 km/hr = <em>x</em> / <em>t</em>

you can instead solve for <em>t</em> in terms of <em>x</em>, then plug that into the distance equation.

<em>t</em> = <em>x</em> / (17.5 km/hr)

then

<em>x</em> ≈ (22.0 km/hr) (<em>x</em> / (17.5 km/hr)) - 7.33 km

<em>x</em> ≈ 1.25714 <em>x</em> - 7.33 km

0.25714<em>x</em> ≈ 7.33 km

<em>x</em> = (7.33 km) / 0.25714 ≈ 28.5 km

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