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Westkost [7]
1 year ago
6

Sage measured her rectangular house as 15.24 m long and 9.1 m wide. Using her calculator, she multiplied the length by the width

to determine that the area is 138.684 m2. What should she do next to express this value in proper significant digits?
Physics
1 answer:
Vitek1552 [10]1 year ago
3 0

The next step for Sage to express the value she multiplied in proper significant digits is to round up the area to the correct number of significant figures.

<h3>What is significant figures?</h3>

Significant figures are digits that is meaningful with respect to the precision of a measurement.

In other words a digit that is nonzero, followed by a nonzero digit, or (for trailing zeroes) justified by the precision of the derivation or measurement.

According to this question, Sage measured her rectangular house as 15.24 m long and 9.1 m wide. Using her calculator, she multiplied the length by the width to determine that the area is 138.684 m².

  • Decimal notation: 138.684
  • No. of significant figures: 6
  • No. of decimals: 3
  • Scientific notation: 1.38684 × 10²

Learn more about significant figures at: brainly.com/question/29153641

#SPJ1

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(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

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v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

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v x 13600 x g = V x 920 x g

v / V = 0.0676

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a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

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a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

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x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

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