Alan

If you are referring to the friction coefficient question then yes. g = 9.81m/s²

You probably mean "degrees" not "degrades", in which case, one way is to enter 179+50/60+14/3600 - (119+58/60+41/3600) to get the answer in degrees, remembering that there are 60 minutes in a degree and 3600 seconds in a degree.  *Edited to correct number of seconds in degree!

$${\mathtt{179}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{50}}}{{\mathtt{60}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{14}}}{{\mathtt{3\,600}}}}{\mathtt{\,-\,}}\left({\mathtt{119}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{58}}}{{\mathtt{60}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{41}}}{{\mathtt{3\,600}}}}\right) = {\mathtt{59.859\: \!166\: \!666\: \!666\: \!666\: \!7}}$$

This gives 59° plus a bit.  Turn the extra into minutes by multiplying by 60:

$${\mathtt{0.859}}{\mathtt{\,\times\,}}{\mathtt{60}} = {\frac{{\mathtt{2\,577}}}{{\mathtt{50}}}} = {\mathtt{51.54}}$$

This is 51' plus a bit.  Turn the extra into seconds by multiplying by another 60.

$${\mathtt{0.54}}{\mathtt{\,\times\,}}{\mathtt{60}} = {\frac{{\mathtt{162}}}{{\mathtt{5}}}} = {\mathtt{32.4}}$$

This is 32'' to the nearest second.

So the overall result is 59°51'32''

The first equation, which should probably be written as 1/2*m*v₀² = F_fr*D, says that the initial kinetic energy is equal to the work done by friction in bringing the object to rest (the right-hand side is frictional force*distance which is work, which is a form of energy transfer).

The frictional force is usually represented by F_fr = μ*F_n, where F_n is the normal force; that is, the force holding the object on the floor, namely, its weight (so F_n = m*g, where m is the object's mass and g is the acceleration of gravity).   μ is called the coefficient of friction.

So, putting this together, we end up with: 1/2*m*v₀² = μ*m*g*D which can be rearranged to find μ = (1/2*m*v₀²)/(m*g*D) and the m cancels so μ = (1/2*v₀²)/(g*D).

Does this make it clearer?

That's interesting kitty!  What browser do you use?

On the main Questions page I see the beginning of the question in words, but it cuts off part way through the first sentence.  Then, when I select the question I see some words interspersed by icon's that represent images. 

When I try to trace what these are by using "inspect element" I see URLs that end in .gif.   Attempting to link to these I get the message that only those registered for the course can see them (or some such message - I don't remember the exact words!).

The images in the question appear to be part of an online course.  They are not available to view for anybody not registered for the course, which is most of us, I guess (presumably kitty<3 is an exception, since she seems to know what's in them).  

Ok, I watched the video.  It was very clear for the most part (though a little puzzling that he said he wasn't going to use Kirchoff's laws and then proceeded to use Kirchoff's current law to set up his equations!  I guess he meant he wasn't going to use both laws together.).

However, if you don't understand how to use matrices, the last part, where he gets the voltages by producing results obtained with a calculator, must seem like black magic!  With just two simultaneous equations it's not too difficult to solve the matrix equations by hand (though a little tedious, which is why he just used a calculator).  It is worth taking the time to understand enough about matrices so that you don't think it's black magic, even if you do eventually use a calculator or computer software to do the number crunching.

As long as you only deal with two simultaneous equations, it doesn't much matter which approach you take.  However, the more simultaneous equations you have to solve, the better it is to adopt a matrix approach (though, inevitably, computer software using advanced numerical solution algorithms are required to do the number crunching for large matrices), so it's worth learning the principles of how to do this with just two. 

Use the "^" character.  e.g. 3^2 will give 9.

From Pythagoras's theorem we have diagonal = √(100² + 50²) feet

$${\mathtt{diagonal}} = {\sqrt{{{\mathtt{100}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{50}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{diagonal}} = {\mathtt{111.803\: \!398\: \!874\: \!989\: \!484\: \!8}}$$ feet

To the nearest foot this is 112 ft