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lesantik [10]
2 years ago
11

What is the difference between accurate data and reproducible data?

Physics
2 answers:
Shtirlitz [24]2 years ago
5 0

Answer:


Accuracy is ‘the state of being correct’ and on the other hand precision is ‘the state of being exact.


Explanation:


Accuracy shows the nearness of the measurement with the actual measurement while precision shows the nearness of an individual measurement with other measurements.


Accuracy focuses on the errors caused by the problem in the instrument. while precision is concerned with errors, which occurs periodically with no recognizable pattern.


If the measurement is precise but inaccurate, the result will not be the expected one. And if the result is accurate but imprecise, then there are huge variations in the measurements. And finally, if the actual measurement is neither accurate nor precise, then the result would not have correctness and exactness at the same time.

vladimir1956 [14]2 years ago
4 0
Accuracy is a general concept while precision is more of a mathematical concept.
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Negative

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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
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Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

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Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

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= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

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