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3 years ago

Add these measurements, using significant digit rules: 1.0090 cm + 0.02 cm = cm

Physics

1 answer:

Luda [366]3 years ago

8 0

Answer: " 1.03 " .

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Explanation:

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When we add "measured" or "obtained values" ; the rules for "signficant digits" (also known as "significant figures" ; or "sig figs") are different from the rules we use when we multiply "measured or obtained values".

We are given 2 (two) "measured or obtained values" ; and we are asked to add these values together using the "significant digit rules.

In this case, we add up the "sum total value" of all the values—Not taking into account the number of significant figures. Then, we take that "sum value" ; and we round that value to the number of "decimal places" that corresponds to the the number of decimals places in the "measured/obtained valued" that occurs in the value using to calculate the sum —the one that also happens to have the least number of significant digits among all the values used to calculate the "sum total".

As such:

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1.0090 cm

+ 0.02 cm

________________

1.0290 cm.

→ Now, there were two (2) values used to calculate the "sum total" ;

1) 1.0090 cm ; which has 5 (five) significant digits.

2) 0.02 cm ; which as 1 (one) significant digits.

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So, " 1" is less than "5" ; so: " 0.02" is the value within the calculations used to find the sum total — that has the "least" number of significant digits.

→ " 0.02 " ; happens to have 2 (two) decimal spaces.

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So, we take the "sum total" :

→ " 1.0290 cm " ;

→ And round this number to: " 2 (two) decimal places:

→ 1.029... ;

→ If the number after the second decimal place is from "1 to 4" ; we round down. If the number after the second decimal place is

from "5 and 9" ; to round up. Since the third decimal place is: "9" ;

[Note: " 1.02<u>9</u>.." ; is between "5 to 9" } ; we round up:

→ The answer is: " 1.03 " .

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Hope this is helpful to you!

Wishing you the best!

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<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

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<h3>Mass of the plane at the position of the rear wheels</h3>

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Let the mass of the plane at rear wheels = M2

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<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

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