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Ierofanga [76]
3 years ago
13

This list shows the heights in feet of the 30 tallest mountains in the United States. 14,269 14,361 14,470 14,831 16,237 14,270

14,370 14,494 14,950 16,390 14,286 14,410 14,500 15,300 16,500 14,294 14,420 14,530 15,638 17,400 14,309 14,421 14,573 15,885 18,008 14,345 14,433 14,730 15,979 20,320 Jennifer is making a frequency table for the mountain heights. She has created 7 intervals of 1000 feet as shown below. U.S. Mountains Height (ft) Tally Frequency 14,000 - 14,999 15,000 - 15,999 16,000 - 16,999 17,000 - 17,999 18,000 - 18,999 19,000 - 19,999 20,000 - 20,999 What are the frequencies for each interval? a. Frequency. 20, 4, 3, 1, 1, 0, 1. c. Frequency: 20, 3, 4, 1, 0, 1, 1. b. Frequency. 19, 5, 3, 1, 1, 0, 1. d. Frequency: 19, 5, 3, 1, 0, 1, 1. Please select the best answer from the choices provided A B C d

Physics
1 answer:
antoniya [11.8K]3 years ago
8 0

Answer: a. Frequency. 20, 4, 3, 1, 1, 0, 1

Explanation:

Interval(Height in feet) - - - - - - - - frequency

14000 - 14999 - - - - - - - - - - - - - - - 20

15000 - 15999 - - - - - - - - - - - - - - - 4

16000 - 16999 - - - - - - - - - - - - - - - 3

17000 - 17999 - - - - - - - - - - - - - - - 1

18000 - 18999 - - - - - - - - - - - - - - - 1

19000 - 19999 - - - - - - - - - - - - - - - 0

20000 - 20999 - - - - - - - - - - - - - - 1

Frequency : (20, 4, 3 1, 1, 0, 1)

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Answer: B

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3 years ago
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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

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3 years ago
Explain in your own words how the Doppler Effect is also applicable in our study of light.
arlik [135]

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Answer:

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Explanation:

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Where:

m=mass\hspace{3}of\hspace{3}the\hspace{3}object

v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval

v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.

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