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3 years ago

Four students measure the mass of a standard mass that has an accepted value of 150.0 g. They post their results in a table.

Physics

1 answer:

Eddi Din [679]3 years ago

4 0

Explanation:

This problem seeks to ensure that the student comprehends the meaning of precision and accuracy.

These two key terms are used to express how reliable experimental values obtained are.

Precision is the ability to consistently reproduce a result.
Accuracy is the nearness of the measure value to the true value.
The different between the measured and true values gives the error in the result.

In the problem given, the most accurate reading would be one that has value close to the standard mass of 150g. Any student that reports this value would be reported as accurate.

The precise reading would be for the student that is able to repeatedly produce the same kind of values. This is irrespective of whether the values are close to the true value or not.

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Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

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Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

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Answer:

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A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m

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Answer:

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Explanation:

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