The most common atom of iron has 26 protons and 30 neutrons in its nucleus. What are its atomic number, atomic mass, and number of electrons if it is electrically neutral? This atom has atomic number 26, atomic mass 56, and has 26 electrons.
<em>Answer:</em>
<h3><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>True</em></h3>
- <em>Because </em><em>Gravity is the force of attraction between two objects, and Earth's gravity pulls matter downward, toward its center. It pulls precipitation down from clouds and pulls water downhill. Gravity also moves air and ocean water. ... Gravity pulls denser air and water downward, forcing less dense air and water to move upward.</em>
<em>Carryonlearning</em>
Fossil fuels like coal, natural gas and oil.
Hopes this helps!
Answer:
1. t = 0.0819s
2. W = 0.25N
3. n = 36
4. y(x , t)= Acos[172x + 2730t]
Explanation:
1) The given equation is

The relationship between velocity and propagation constant is

v = 15.87m/s
Time taken, 

t = 0.0819s
2)
The velocity of transverse wave is given by


mass of string is calculated thus
mg = 0.0125N

m = 0.00128kg


0.25N
3)
The propagation constant k is

hence

0.036 m
No of wavelengths, n is

n = 36
4)
The equation of wave travelling down the string is
![y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]](https://tex.z-dn.net/?f=y%28x%2C%20t%29%3DAcos%5Bkx%20-wt%5D%5C%5C%5C%5Cbecomes%5C%5C%5C%5Cy%28x%20%2C%20t%29%3D%20Acos%5B%28172%20rad.m%29x%20%2B%20%282730%20rad.s%29t%5D)
![without, unit\\\\y(x , t)= Acos[172x + 2730t]](https://tex.z-dn.net/?f=without%2C%20unit%5C%5C%5C%5Cy%28x%20%2C%20t%29%3D%20Acos%5B172x%20%2B%202730t%5D)