<span>Radio Waves, Microwaves, infra-Red, Visible spectrum, Ultraviolet radiation, x-rays, Gamma Rays. Then again I could be wrong.</span>
Answer:
Hello, how's your day going?
if humanity came together and made a base on the moon, it would be revolutionary. The point of a base on the moon would have multiple purposes. for example, some think that the moon contains valuable metals such as iron and titanium. a base would serve as a place for workers harvesting metals to rest. Obviously or not most of the iron harvesting would be done automatically by robots and such.
If such a base were constructed on the moon, it would be the begining of people living on other worlds and would be a great start for a base on Mars.
Hope it helped
Spiky Bob
Formulas you need for this problem:
F= mass•acceleration
KE= mass•velocity^/2
Acceleration= final velocity-intial velocity/time
time= distance/speed
t= 97.5\10= 9.75 seconds
Acc= 10-20/9.75= -1.03 m/s/s
These are the answers :)
F=650•-1.03= -669.5N
KE= 650•100/2= 32,500J or 32.5KJ
Answer : The mechanical energy of the rock is, 134 J
Solution :
As we know that the mechanical energy is equal to the sum of the kinetic energy and the potential energy.
The formula used for mechanical energy is,

where,
M.E = mechanical energy
K.E = kinetic energy
P.E = potential energy
m = mass of an object = 2 Kg
v = velocity of an object = 6 m/s
h = height = 5 m
g = acceleration due to gravity = 
Now put all the given values in the above formula, we get the mechanical energy of the rock.
![M.E=[\frac{1}{2}(2Kg)\times (6m/s)^2]+(2Kg\times 9.8m/s^2\times 5m)=134\text{ Kg }m^2s^{-2}=134J](https://tex.z-dn.net/?f=M.E%3D%5B%5Cfrac%7B1%7D%7B2%7D%282Kg%29%5Ctimes%20%286m%2Fs%29%5E2%5D%2B%282Kg%5Ctimes%209.8m%2Fs%5E2%5Ctimes%205m%29%3D134%5Ctext%7B%20Kg%20%7Dm%5E2s%5E%7B-2%7D%3D134J)
Therefore, the mechanical energy of the rock is, 134 J
Answer:
the natural length of the spring is 9 cm
Explanation:
let the natural length of the spring = L
For each of the work done, we set up an integral equation;
![5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)](https://tex.z-dn.net/?f=5.4%20%3D%20%5Cint%5Climits%5E%7B21-l%7D_%7B15-l%7D%20%7Bkx%7D%20%5C%2C%20dx%20%5C%5C%5C%5C5.4%20%3D%20%5B%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%5D%5E%7B21-l%7D_%7B15-l%7D%5C%5C%5C%5C5.4%20%3D%20%5Cfrac%7Bk%7D%7B2%7D%20%5B%2821-l%29%5E2%20-%20%2815-l%29%5E2%5D%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B2%285.4%29%7D%7B%2821-l%29%5E2%20-%20%2815-l%29%5E2%7D%20%20%5C%20%5C%20%5C%20-----%281%29)
The second equation of work done is set up as follows;
![9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)](https://tex.z-dn.net/?f=9%20%3D%20%5Cint%5Climits%5E%7B27-l%7D_%7B21-l%7D%20%7Bkx%7D%20%5C%2C%20dx%20%5C%5C%5C%5C9%20%3D%20%5B%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%5D%5E%7B27-l%7D_%7B21-l%7D%5C%5C%5C%5C9%20%3D%20%5Cfrac%7Bk%7D%7B2%7D%20%5B%2827-l%29%5E2%20-%20%2821-l%29%5E2%5D%20%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B2%289%29%7D%7B%2827-l%29%5E2%20-%20%2821-l%29%5E2%7D%20%5C%20%5C%20%5C%20-----%282%29)
solve equation (1) and equation (2) together;


Therefore, the natural length of the spring is 9 cm