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shutvik [7]
3 years ago
9

A student measures the mass of a 1.0 kg standard bar. He obtains measurements of 0.77 kg, 0.78 kg, and 0.79 kg. Which describes

his measurements
Physics
1 answer:
Vadim26 [7]3 years ago
4 0

Complete question

A student measures the mass of a 1.0 kg standard bar. He obtains measurements of 0.77 kg, 0.78 kg, and 0.79 kg. Which describes his measurements

a)precise but not accurate

b)accurate but not precise

c)neither precise nor accurate

d)both precise and accurate

Answer:

The measurement is precise but not accurate

Explanation:

A measurement can either be precise or accurate.

  • A Precise measurement describes how close the measured values are to each other.
  • An accurate measurement describes how close a measured value is to the true value.

In this question, the measured values (0.77 kg, 0.78 kg and 0.79 kg) are far from the true value (1.0 kg), therefore the measurement is not accurate.

However, the measured values (0.77 kg, 0.78 kg and 0.79 kg) are close to each other, therefore the measurement is precise.

Therefore the correct option is 'a' the measurement is precise but not accurate

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B. The time taken to slide until he stops is 1.2 s

<h3>How to determine the force of friction</h3>
  • Mass (m) = 81.5 Kg
  • Coefficient of friction (μ) = 0.7
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
  • Frictional force (F) =?

F = μN

F = 0.7 × 798.7

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<h3>A. How to determine the acceleration</h3>
  • Mass (m) = 81.5 Kg
  • Frictional force (F) = 559.09 N
  • Acceleration (a) =?

a = F / m

a = 559.09 / 81.5

a = 6.86 m/s²

<h3>B. How to determine the time </h3>
  • Initial velocity (u) = 8.23 m/s
  • Final velocity (v) = 0 m/s
  • Decceleration (a) = -6.86 m/s²
  • Time (t) =?

a = (v – u) / t

t = (v – u) / a

t = (0 – 8.23) / -6.86

t = 1.2 s

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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