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SCORPION-xisa [38]
2 years ago
10

An atom of argon has a radius of 71.pm and the average orbital speed of the electrons in it is about ×3.9107/ms. calculate the l

east possible uncertainty in a measurement of the speed of an electron in an atom of argon. write your answer as a percentage of the average speed, and round it to 2 significant digits.
Physics
1 answer:
Anna11 [10]2 years ago
7 0

Answer: 2.1 %

Explanation:

The radius of the Argon atom, r = 71 pm = 7.1 × 10 ⁻¹¹ m

Average orbital speed of electrons, v = 3.9 × 10⁷ m/s

From uncertainty principle:

Δx m Δv ≥ h/4π

mass of electron, m = 9.1 ×10⁻³¹ kg

Δx = radius of the argon atom = 7.1 × 10 ⁻¹¹ m

\Rightarrow \Delta v = \frac {6.626 \times 10^{-34} m^2kg/s}{4\times 3.14 \times 7.1 \times 10^{-11} m \times 9.1 \times 10^{-31} kg}

\Delta v = 8.2 \times 10^5 m/s

Percentage uncertainty:

\frac{\Delta v}{v} \times 100\% = \frac {8.2 \times 10^5 m/s}{3.9 \times 10^7 m/s} \times 100 \%= 2.1 \%

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Answer:

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Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

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The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j  

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0   since i × i = 0, j × j = 0, i × j = k and j × i = -k

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2 years ago
You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly
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A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

<span>The reading will drop to 0 instantly</span>

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A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

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Length of the string, l = r = 2 ft

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F=\dfrac{mv^2}{r}-mg

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F=\dfrac{mv^2}{r}+mg

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(c) Let h is the height where the ball at certain time from the top. So,

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Answer:

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