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Bumek [7]
3 years ago
14

If different groups of scientist have access to the same data, how can they draw different conclusions?

Physics
2 answers:
kolbaska11 [484]3 years ago
8 0

Answer:

This may depend on the model in which the scientist is working with, for example, different models see the electrons in different ways, where ones like the Drude model are very simplistic and other ones, like the used for medical physics are a little bit more complex. So for example for the same data, a given number of electrons in a given material, the different groups of scientist can draw different results because they may be trying different experiments in which they use a different models to "think" the problem that they want to solve, where ones may want to see how the material conducts heat (Drude model) and others may want to see how the material can be used in a radiating machine (more complex models)

vazorg [7]3 years ago
6 0

If a group of scientists have access to one data, from the data they can draw conclusions either through mathematics or just thought experiments.

Those thought experiments is different for any scientist, no one thinks the same especially when the topic is difficult.

For example when talking about parallel universes, scientists have come up with the weirdest examples of a multiverse. Some thinking of a brane universe, while others say that its a landscape universe, quilted universe. All of their 'evidence' seems correct but they have opposite meanings.

A weird analogy is 'religion'. All the religions seem to have 'evidences' (hardly) that attract people towards it, they all make sense but that doesn't mean that their evidence is right.

----

Now if they're trying to break down the data using maths, there could be a great uncertainty and measurement error that if done enough could change the whole idea behind the data.

Interesting question, I can babble for days for this but lets keep it as that

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7 0
3 years ago
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum
Alina [70]

Answer:

  • Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
  • Uncertainty in momentum (∆P) = ?
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\longrightarrow \:  \:  \sf\Delta x .\Delta p =  \dfrac{h}{4\pi}

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} {4 \times  \frac{22}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} { \frac{88}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34} \times 7} { 8 }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 8  \times 24 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 2}  \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 21}   } { 192}

\longrightarrow \:  \:  \sf\Delta p = 22.822\times  {10}^{ - 21}

\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

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