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sp2606 [1]
3 years ago
13

Mary travelled 70mph due north

Physics
1 answer:
tiny-mole [99]3 years ago
3 0
Velocity because It is defined as the change in the position with respect to the time. Velocity is a vector quantity that means it depends on the magnitude and direction of an object. The S.I unit of velocity is, m/s

Acceleration : It is defined as the rate of change of velocity of an object wit respect to the time. Acceleration is a vector quantity that means it depends on the magnitude and direction of an object. The S.I unit of acceleration is, m/s to the power of 2

Distance : It is defined as the how far an object has traveled in time. Distance is a scalar quantity that means it is depends on the magnitude of an object only. The S.I unit of distance is, m

Speed : It is defined as the distance traveled by an object in unit time. Speed is also a scalar quantity. The S.I unit of speed is, m/s

Mary traveled 70 miles/hour due north. This is an example of velocity. 70 miles/hour tell us about the magnitude of the object and north tell us about the direction of an object.
Hence, the correct option is, 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
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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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Answer:

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