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VikaD [51]
3 years ago
14

An element's atomic number is 59. How many electrons would an atom of this element have?

Physics
1 answer:
katrin [286]3 years ago
6 0
Hi,
An element’s atomic number tells how many protons an atom of this element has. Protons have a positive electric charge, while electrons have a negative one. If an atom is electrically neutral (as in this case, otherwise it should have been specified) the number of protons and electrons are equal. So, an atom of the element would have 59 electrons.

Hope this helps! If my answer was not clear enough or you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistake.
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2 years ago
a sprinter starts from rest and accelerates at a rate of 0.16 ms ^-2 over a distance of 50.0 meters . How fast is the athlete tr
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Answer:

Final velocity v = 4 m/s

Explanation:

Given:

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3 years ago
Which data set has the largest standard deviation
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Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}

Where x_i is the value of each measurement, n is the number of elements in the set, and \mu is the average or media of the values, defined as

\displaystyle \mu=\frac{\sum x_i}{n}

Let's analyze each set of data:

A.3,4,3,4,3,4,3

The average is

\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43

Computing the stardard deviation:

\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}

\sigma=0.5

B.1,6,3,15,4,12,8

The average is

\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7

Computing the stardard deviation:

\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}

\sigma=4.7

C. 20, 21,23,19,19,20,20

The average is

\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29

Computing the stardard deviation:

\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}

\sigma=1.3

D.12,14,13,14,12,13,12

The average is

\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86

Computing the stardard deviation:

\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}

\sigma=0.8

We can see the data set marked as B has the largest standard deviation

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