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Softa [21]
3 years ago
12

When is zero considered significant​

Physics
2 answers:
Semenov [28]3 years ago
7 0

Answer:

The trailing zero to the right in the decimal place and the zero present between the non zero digit are significant.

Explanation:

The rules which describes about the significant digits are discussed below as:

  • All the digits which are non zero are significant. Example 178 has three significant digits.
  • The zero which is present in between two non zero digits are consider significant. Example 507 has 3 significant digits.
  • The leading zeroes are not consider significant. Example 0.00676 has 3 significant digits only.
  • Trailing zero to the right in the decimal is consider significant. Example 3.0 has 2 significant digits.
  • Trailing zero which is present in the whole number without any decimal are not significant. Example 9800 has only 2 significant digit.

vazorg [7]3 years ago
3 0

If a zero is found between significant digits, it is significant. Zeros can be used as (insignificant) place holders to the left of significant digits if the number is a decimal. For example, a mass of 42 g has two significant digits. Expressed in kilograms, the mass of 0.042kg should still have two significant digits.

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Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

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Which molecule will undergo only london dispersion forces when interacting with other molecules of the same kind?
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Explanation:

London dispersion forces will form between non-polar molecules(polar ) that are symmetrical  like O₂, H₂, Cl₂ and noble gases.

  • The attraction here is because non-polar molecules becomes polar due to the constant motion of its electrons.
  • This lead to an uneven charge distribution at an instant.
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Learn more:

Intermolecular forces brainly.com/question/10602513

#learnwithBrainly

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