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pshichka [43]
3 years ago
15

A marathon runner completes a 42.188–km course in 2 h, 36 min, and 12 s. there is an uncertainty of 23 m in the distance travele

d and an uncertainty of 3 s in the elapsed time. calculate the percent uncertainty in the distance.
Physics
1 answer:
timurjin [86]3 years ago
4 0

The percentage uncertainty in the distance is <u>0.054%</u>.

Uncertainty in a measurement measures the deviation of the measured values from the true value.

The distance d can be expressed in the form <em>d+/-Δd, </em>where <em>Δd i</em>s the absolute uncertainty in the measurement of distance.

The percentage uncertainty  is given by,

Percentage uncertainty=\frac{\Delta d}{d} *100

The uncertainty in the measurement of distance is 23 m. Express the uncertainty in km.

\Delta d =\frac{23 m}{1000 m/km} =0.023 km

Calculate the percentage uncertainty in the distance.

\frac{\Delta d}{d} *100 =\frac{0.023 km}{42.188 km} *100=0.054 %

Thus, the percentage uncertainty in the measurement of distance is <u>0.054%</u>

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Answer:

<h3>The answer is 0.54 g/cm³</h3>

Explanation:

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density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{10.5}{19.3}  \\  = 0.54404145...

We have the final answer as

<h3>0.54 g/cm³</h3>

Hope this helps you

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Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the
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To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

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L = 5.8cm = 5.8*10^{-2}m

A = 1.5cm^2 = 1.5*10^{-4}m^2

The number of the atoms N, can be calculated as,

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N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)

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Then applying the equation about the dipole moment associated with an iron bar we have,

\mu = \alpha *N

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\tau = \mu B sin(90)

\tau = (20.72)(2.2)

\tau = 45.584N.m

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

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