Answer:This is what i could solve hope it helps :)
Number17
[tex]H = \frac{m(v^2-u^2}{2gx} \\Cross - multiply\\2Hgx = m (v^2-u^2)\\2Hgx = mv^2-mu^2\\\frac{2Hgx+mu^2}{m} = \frac{mv^2}{m} \\Divide -both -sides-of-the-equation-by-m\\v^2 = \frac{2Hgx+mu^2}{m}\\Square - root-both -sides -to -eliminate the square\\v =\sqrt{\frac{2Hgx+mu^2}{m}}[/tex]
Step-by-step explanation:
Number 19
[tex]T =2\pi \sqrt{\frac{1}{MH}}\\Divide - both - sides-of-the-equation-by 2\pi\\\frac{2\pi \sqrt{\frac{1}{MH}}}{2\pi }=\frac{T}{2\pi }\\\sqrt{\frac{1}{MH}}=\frac{T}{2\pi }\\SQUARE-BOTH -SIDES\\\left(\sqrt{\frac{1}{MH}}\right)^2=\left(\frac{T}{2\pi }\right)^2\\\frac{1}{MH}=\frac{T^2}{4\pi ^2}\\CROSS-MULTIPLY\\MHT^2={4\pi ^2\\\\DIVIDE BOTH SIDES OF THE EQUATION BY HT^2\\M=\frac{4\pi ^2}{HT^2}[/tex]
Number 20 &21
[tex]A =\frac{1}{2} m(v^2-u^2)\\Divide both sides by \frac{1}{2} m\\\frac{\frac{1}{2}m\left(v^2-u^2\right)}{\frac{1}{2}m}=\frac{A}{\frac{1}{2}m}\\2Am = v^2-u^2\\Move- v^2- to- the -other -side- to- isolate- u\\\\u^2 = 2Am-v^2\\Square-root-both-sides-to -eliminate-the-square-on-u.\\u = \sqrt{2Am-v^2}[/tex]
