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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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A) The work done by the electric field is zero

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C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

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\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

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This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

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B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

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cos 0^{\circ}=1

Therefore, the work done by the electric force is

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F=qE is the magnitude of the electric force. Since

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q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

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Answer:

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Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

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x = \frac {-49*0.002*9.8}{24}

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