Alan

Just divide the mass by Avogadro's number (given in the question as 6*10²³)

$$\frac{0.197}{6\times 10^{23}}= 0.03283\times 10^{-23}kg$$

or 

$$3.28\times 10^{-25}kg$$

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This refers to the length of a column of mercury used to measure pressure (or pressure drop).

Pressure drop of a liquid can be expressed as Δp = ρgh where ρ is density, g is gravitational acceleration and h is the height of the column of liquid.  Sometimes, just the value of h is given (though this is a rather dated practice these days) and is referred to as a "head" of pressure.  With mercury, the head is usually given in centimetres or millimetres, but if water is the liquid it was often given in feet.  Thus, for water at least, "heads" were measured in feet!

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I suspect this is all that is being asked for:

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1. Change each number to an improper fraction

2 2/7 = 16/7,    3 1/3 = 10/3

2. To divide by a fraction turn the fraction in the denominator upside down and multiply instead:

(16/7)/(10/3) =  (16/7)*(3/10)

$$\left({\frac{{\mathtt{16}}}{{\mathtt{7}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{10}}}}\right) = {\frac{{\mathtt{24}}}{{\mathtt{35}}}} = {\mathtt{0.685\: \!714\: \!285\: \!714\: \!285\: \!7}}$$ 

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Assuming no resistance from the air (unrealistic in the case of a football!) the time of flight is given by:

0 = 19.5*sin(45°)*t - (1/2)*9.8*t², taking g = 9.8m/s² (this comes from considering the vertical motion)

$${\mathtt{t}} = {\frac{{\mathtt{19.5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{9.8}}}} \Rightarrow {\mathtt{t}} = {\mathtt{2.813\: \!996\: \!374\: \!111\: \!530\: \!6}}$$ seconds

The horizontal distance travelled by the ball is  s = 19.5*cos(45°)*t

$${\mathtt{s}} = {\mathtt{19.5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{2.813\: \!996\: \!374\: \!111\: \!530\: \!6}} \Rightarrow {\mathtt{s}} = {\mathtt{38.801\: \!020\: \!408\: \!212\: \!922\: \!5}}$$ metres

The player must therefore run a distance of (55 - s) metres in a time t seconds, so his average speed must be (55 - s)/t m/s

$${\mathtt{v}} = {\frac{\left({\mathtt{55}}{\mathtt{\,-\,}}{\mathtt{38.801}}\right)}{{\mathtt{2.814}}}} \Rightarrow {\mathtt{v}} = {\mathtt{5.756\: \!574\: \!271\: \!499\: \!644\: \!6}}$$ m/s

or v ≈ 5.76 m/s

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Not if n is an integer:

5*2ⁿ = 200 means 2ⁿ = 40 , but 2⁵ = 32 and 2⁶ = 64.

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Mass is the same wherever you are.

Weight is mass*g, so on Earth weight = 50*9.81 = 490.5 N, and on the Moon weight = 490.5/6 = 81.75 N

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Sure it has an answer!  You can see by drawing a graph:

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Another approach to this is to divide through by the product on the right-hand side of each equation, so that we get:

1/b + 1/a = 4

1/c + 1/b = 5

1/a + 1/c = 3

Now let A = 1/a, B = 1/b and C = 1/c, so the equations become

B + A = 4            (1)

C + B = 5            (2)

A + C = 3            (3)

Subtract (3) from (2) to get B - A = 2

Add this to (1) to get 2B = 6 so B = 3, so b = 1/3

Put B back into (1) to get A = 1, so a = 1

Put B back into (2) to get C = 2, so c  = 1/2

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Since this equation is easy to solve using the quadratic formula (see below), what is your criterion for "approximate"?  Is it just a number of decimal places, or are you expected to guess a few values and home in on the results, or what?

$${\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{58}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{7}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{58}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{7}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.659\: \!396\: \!157\: \!980\: \!558\: \!3}}\\
{\mathtt{x}} = {\mathtt{1.516\: \!539\: \!015\: \!123\: \!415\: \!5}}\\
\end{array} \right\}$$

To two decimal places the values are x = -0.66 and x = 1.52

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