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3 years ago

Which answer is a scientifically accurate description of velocity?

1 answer:

4 0

The boat traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

Explanation:

Velocity is the rate of change of displacement with time. It is a physical quantity that has both magnitude and direction.

When expressing velocity, we must consider the directional attribute as it distinguishes it from speed.

Statements:

The hiker began hiking the park's five-mile north hiking trail at 8:00 a.m. with plenty of daylight left to complete the hike before dark.

We are given the displacement of the hiker but the duration of traveling was not indicated. The starting time is known and there is nothing we can infer from it

The motorcyclist traveled along a highway at 55 mph and reached the campsite in less than the two-hour estimated time.

55mph is the speed of the motorcyclist and not velocity. There was no directional attribute.

Walking south along the two-mile canyon trail generally takes less time than traveling the same trail north because walking south is all downhill

This is not quantifiable.

The boat traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

The 200m marker north is the displacement and the time is 5minutes. This is the most scientifically accurate description of velocity.

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The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an

yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

$\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}$

$\frac{1}{f}=0.028$

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

$p=\frac{1}{f}$

Here, p is the power of the lens.

Put f= 35.71 cm.

$p=\frac{1}{35.71}$

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

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