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vodomira [7]
3 years ago
15

Four students measure the mass of a standard mass that has an accepted value of 150.0 g. They post their results in a table.

Physics
1 answer:
Eddi Din [679]3 years ago
4 0

Explanation:

This problem seeks to ensure that the student comprehends the meaning of precision and accuracy.

These two key terms are used to express how reliable experimental values obtained are.

  • Precision is the ability to consistently reproduce a result.
  • Accuracy is the nearness of the measure value to the true value.
  • The different between the measured and true values gives the error in the result.

In the problem given, the most accurate reading would be one that has value close to the standard mass of 150g. Any student that reports this value would be reported as accurate.

The precise reading would be for the student that is able to repeatedly produce the same kind of values. This is irrespective of whether the values are close to the true value or not.

learn more:

Statistics brainly.com/question/4401748

#learnwithBrainly

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Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
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Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

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    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

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    y₂ = 0 + ½ g (t-t₀)²

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Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

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    S = ½ g t²– ½ g (t-t₀)²

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This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

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Explanation:

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An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density ρ.
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** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **

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