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tresset_1 [31]
3 years ago
15

If a measurement is 4.2 cm and its possible error is 0.04 cm, the relative error is?​

Physics
1 answer:
galben [10]3 years ago
3 0

Given the value of possible error, the relative error of the measurement is 0.0095.

Given the data in the question;

  • Measurement taken; m = 4.2cm
  • Absolute or possible erorr; e = 0.04m
  • Relative error; RE = \ ?

<h3>Relative Error</h3>

Relative error is simply the ratio of the absolute or possible error of a measurement to the measurement its self.

Relative error is expressed as;

RE = \frac{e}{m}

Where e is the absolute or possible error and m is the measurement being taken.

To determine the relative error, we substitute our given values into the expression above.

RE = \frac{0.04cm}{4.2cm} \\\\RE = 0.0095%

Therefore, given the possible error, the relative error of the measurement is 0.0095.

Learn more about relative error: brainly.com/question/17016021

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Applying Newton's Second Law of Motion, the acceleration of the ball is 16.8 m/s^2

<u>Given the following data:</u>

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To find ball's acceleration, we would apply Newton's Second Law of Motion:

First of all, we would determine the net force acting on the ball.

Net \; force = Upward\;force + Downward\;force

Downward\;force = 0.5 × 9.8

Downward force =  4.9 N

Net \; force = 3.5 + 4.9

Net force = 8.4 N

Mathematically, Newton's Second Law of Motion is given by this formula;

Acceleration = \frac{Net\;force}{Mass}\\\\Acceleration = \frac{8.4}{0.5}

<em>Acceleration = 16.8 </em>m/s^2<em />

Therefore, the acceleration of the ball is 16.8 m/s^2

Read more here: brainly.com/question/24029674

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What happens to jetstream’s as they get closer to the equator
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<u>ACID</u>

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<u>BASE</u>

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A 30.9 kg rocket has an engine
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Answer:

acceleration = 15.8 m/s^2

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