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laila [671]
4 years ago
8

If the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity? - The

accuracy in measuring its velocity becomes uncertain.
- The accuracy in measuring its velocity remains the same.
- The accuracy in measuring its velocity also increases.
- The accuracy in measuring its velocity decreases.
Physics
1 answer:
timofeeve [1]4 years ago
5 0

Answer:

- The accuracy in measuring its velocity decreases.

Explanation:

This can be explained by Heisenberg's uncertainty principle which states that the position and velocity of a particle can be determined together exactly in reality.

This principle, unlike Newtonian mechanics deal with particles at microscopic level like that of an electron where if the accuracy in measurement of particle's position increases there will be  decreased accuracy in measurement of velocity of that particle.

There will be an uncertainty in accuracy in the measurement of particle's position and its velocity.

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3 years ago
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A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle
KATRIN_1 [288]

Answer:

electric flux through the three side = 2.35 N m²/C

Explanation:

given,

equilateral triangle of base = 25 cm

electric field strength = 260 N/C

Area of triangle = \dfrac{\sqrt{3}}{4}a^2

                          =  \dfrac{\sqrt{3}}{4} 0.25^2

                          = 0.0271 m³

electric flux = E. A

                   = 260 × 0.0271

                   = 7.046 N m²/C

since, tetrahedron does not enclose any charge so, net flux through tetrahedron is zero.

electric flux through the three side = (electric flux through base)/3

                               = \dfrac{7.046}{3}

electric flux through the three side = 2.35 N m²/C

4 0
3 years ago
A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potentia
Llana [10]

Answer:

42.3 MV

Explanation:

d = diameter of the metal sphere = 2.15 m

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diameter of the metal sphere is given as

d = 2r

2.15 = 2 r

r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

Potential near the surface is given as

V = \frac{kQ}{r}

V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}

V = 4.23 x 10⁷ volts

V = 42.3 MV

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4 years ago
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<span>a solution is a liquid mixture  </span>
4 0
4 years ago
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This question involves the concepts of work done, pressure, and temperature.

The work done per mol of the air is "-724.71 J/mol".

Using the given formula for work done:

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where,

W = work done per mol = ?

R = universal gas constant = 8.314 J/mol.k

T = absolute temperature = 30°C + 273 = 303 k

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Therefore,

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<u></u>

Learn more about pressure here:

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