Alan

Assuming there is no static friction (!) then the resistive force of the table on the upper mass is 0.6*5*9.8 Newtons.

The force on the lower mass is 5*9.8 Newtons.

The net force on the system is therefore 5*9.8 - 0.6*5*9.8 Newtons, or 0.4*5*9.8 Newtons.

The total mass of the system is 10kg, so, from Newton's second law of motion, the acceleration is net force/mass:

$${\mathtt{acceleration}} = {\frac{{\mathtt{0.4}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{{\mathtt{10}}}} \Rightarrow {\mathtt{acceleration}} = {\mathtt{1.96}}$$

acceleration = 1.96 m/s²

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A mole of shoe strings would be Avogadr's number of shoe strings, namely, 6.02214129x10²³ shoe strings.

The total length of this lot stretched end to end would be 0.00016*6.02214129x10²³ km long.

The distance from Earth to Moon is 384400 km, so the number of times the shoe strings would stretch between them is:

$${\frac{{\mathtt{0.000\: \!16}}{\mathtt{\,\times\,}}{\mathtt{6.022\: \!141\: \!29}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{23}}}}{{\mathtt{384\,400}}}} = {\mathtt{250\,661\,448\,074\,921.956\: \!295\: \!525\: \!494\: \!276\: \!8}}$$

or about 251 trillion times!

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Depends what j is.  If it is √-1 (electrical engineers usually use j instead of i, as they use i for current) then it simplifies to (1 +j)ⁿ

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Yes, this is correct (though a little difficult to read!).

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The product of $$\sqrt{a}$$ and $$a^{\frac{1}{2}}$$ is $$a^{\frac{1}{2}}\times a^{\frac{1}{2}}=a^{\frac{1}{2}+\frac{1}{2}}=a^1=a$$

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The general rule here is $$a^n\times a^k=a^{n+k}$$

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As x gets very large ln( (x+1)/(x-2) ) → ln( x/x ) → ln(1) = 0

and 1/√(x² - 4) → 1/√x² → 1/x → 0

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Do you mean the following:

Let 1.05 be (1 + 0.05) and 9.5 be (10 - 0.5) so 1.05*9.5 = (1 + 0.05)*(10 - 0.5)

This is 1*10 -1*0.5 +0.05*10 - 0.05*0.5 or 10 - 0.5 + 0.5 - 0.025 or 10 - 0.025 or 9.975

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Let the radius of the circle be r.

The diameter of the circle is then 2r, and this is the length of the diagonal of the inscribed square.  This means the length of a side of the inscribed square is 2r/√2, so the area of the inscribed square is (2r/√2)² = 2r².

The diameter is also the length of a side of the circumscribed square, so the area of the circumscribed square is (2r)² = 4r².

Hence the ratio of the area of the inscribed square to that of the circumscribed square is 2r²/4r² = 1/2

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