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In the experiment, the motion of a body was investigated ,as a result of which the average speed and its absolute error were obtained bar (V)=312.0m/s Delta bar (V)=20.483ldots m/s Please indicate the correct record of the final result of this experiment V=(312.0pm 20.5)m/s V=(312.000pm 20.483)m/s V=(312pm 21)m/s V=(312.00pm 20.48)m/s V=(312.0-21)m/s V=(312.0+20.5)m/s

Вопрос

In the experiment, the motion of a body was investigated ,as a result of which the
average speed and its absolute error were obtained
bar (V)=312.0m/s
Delta bar (V)=20.483ldots m/s
Please indicate the correct record of the final result of this experiment
V=(312.0pm 20.5)m/s
V=(312.000pm 20.483)m/s
V=(312pm 21)m/s
V=(312.00pm 20.48)m/s
V=(312.0-21)m/s
V=(312.0+20.5)m/s

In the experiment, the motion of a body was investigated ,as a result of which the average speed and its absolute error were obtained bar (V)=312.0m/s Delta bar (V)=20.483ldots m/s Please indicate the correct record of the final result of this experiment V=(312.0pm 20.5)m/s V=(312.000pm 20.483)m/s V=(312pm 21)m/s V=(312.00pm 20.48)m/s V=(312.0-21)m/s V=(312.0+20.5)m/s

Решения

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мастер · Репетитор 5 лет

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The correct record of the final result of this experiment is:<br /><br />$V=(312.0\pm 20.5)m/s$<br /><br />Explanation: The average speed is given as $\bar{V}=312.0m/s$, and the absolute error is given as $\Delta\bar{V}=20.483\ldots m/s$. When recording the final result, the absolute error should be rounded to one decimal place, which gives $\Delta\bar{V}=20.5m/s$. Therefore, the correct record of the final result is $V=(312.0\pm 20.5)m/s$.
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