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JulijaS [17]
3 years ago
15

"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

yepiece whose focal length is 2.7 cm
Physics
1 answer:
notka56 [123]3 years ago
4 0

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

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A marathon runner completes a 42.238 km course in 2 h, 31 min, and 46 s . There is an uncertainty of 29 m in the distance run an
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Answer:

The percentage uncertainty in the average speed is 0.10% (2 sig. fig.)

Explanation:

Consider the formula for average speed \bar{v}.

\displaystyle \bar{v} = \frac{s}{t},

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The percentage uncertainty of a fraction is the sum of percentage uncertainties in

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What are the percentage uncertainties in s and t in this question?

The unit of the absolute uncertainty in s is meters. Thus, convert the unit of s to meters:

s = \rm 42.238\;km = 42.238\times 10^{3}\;m.

\begin{aligned}\displaystyle \text{Percentage Uncertainty in }s &= \frac{\text{Absolute Uncertainty in } s}{\text{Measured Value of }s}\times 100\% \\ &=\rm\frac{29\; m}{42.238\times 10^{3}\;m}\times 100\%\\ &= 0.0687\%\end{aligned}.

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t = \rm 2\times 3600 + 31\times 60 + 46 = 9106\;s

Similarly,

\begin{aligned}\displaystyle \rm \text{Percentage Uncertainty in }t &= \frac{\text{Absolute Uncertainty in }t}{\text{Measured Value of }t}\times 100\% \\ &=\rm\frac{46\; s}{9106\;s}\times 100\%\\ &= 0.0329\%\end{aligned}.

The average speed \bar{v} here is a fraction of s and t. Both s and t come with uncertainty. The percentage uncertainty in \bar{v} will be the sum of percentage uncertainties in s and t. That is:

\text{Percentage Uncertainty in }\bar{v}\\=(\text{Percentage Uncertainty in } s) + (\text{Percentage Uncertainty in } t)\\ = 0.0687\% + 0.0329\%\\ = 0.010\%.

Generally, keep

  • two significant figures for percentage uncertainties that are less than 2%, and
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The percentage uncertainty in \bar{v} here is less than 2%. Thus, keep two significant figures. However, keep more significant figures than that in calculations to make sure that the final result is accurate.

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Answer:

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f=\frac{3*10^{8}m/s.}{0.365m}

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