Question

Please solve .... it's emergency


The area of cross section of pump plunger a hydraulic machine are 0.04m^2 and 16m^2. Calculate the

1. force acting on Pump plunger if the Hydraulic press overcome the load of 1200kgf


2. If the MA (load/effort) of the handle is 10, calculate the force applied at the end of the pump plunger( give values in kgf)


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Answer 1

$GIVEN :-</p><p></p><p>Perimeter of isosceles triangle = 30 cm.</p><p></p><p>Equal side = 12 cm.</p><p></p><p>TO FIND :-</p><p></p><p>The area of isosceles triangle.</p><p></p><p>SOLUTION :-</p><p></p><p>\begin{gathered}\\ : \implies \displaystyle \sf \: Perimeter \: of \: \triangle = a + b + c \\\end{gathered}:⟹Perimeterof△=a+b+c</p><p></p><p>a = 12 cm.</p><p></p><p>b = 12 cm.</p><p></p><p>c = ?</p><p></p><p>\begin{gathered}\\ : \implies \displaystyle \sf \:30 = 12 + 12 + c \\ \\ \\\end{gathered}:⟹30=12+12+c</p><p></p><p>\begin{gathered}\\ : \implies \displaystyle \sf \:c = 30 - 24 \\ \\ \\\end{gathered}:⟹c=30−24</p><p></p><p>\begin{gathered}\\ : \implies \underline{ \boxed{\displaystyle \sf \:c = 6 \: cm}} \\ \\\end{gathered}:⟹c=6cm</p><p></p><p>____________________</p><p></p><p>\begin{gathered}\\ \dashrightarrow\displaystyle \sf Semi - Perimeter \: of \: \triangle = \frac{a + b + c}{2} \\ \\ \\\end{gathered}⇢Semi−Perimeterof△=2a+b+c</p><p></p><p>\begin{gathered}\dashrightarrow\displaystyle \sf Semi - Perimeter \: of \: \triangle = \frac{12 + 12 + 6}{2} \\ \\ \\\end{gathered}⇢Semi−Perimeterof△=212+12+6</p><p></p><p>\begin{gathered}\dashrightarrow\displaystyle \sf Semi - Perimeter \: of \: \triangle = \frac{30}{2} \\ \\ \\\end{gathered}⇢Semi−Perimeterof△=230</p><p></p><p>\begin{gathered}\dashrightarrow \underline{ \boxed{\displaystyle \sf Semi - Perimeter \: of \: \triangle = 15 \: cm}} \\ \\\end{gathered}⇢Semi−Perimeterof△=15cm</p><p></p><p>___________________</p><p></p><p>\begin{gathered}\\ \longmapsto\displaystyle \sf Area \: of \: \triangle = \sqrt{s(s - a)(s - b)(s - c)} \\\end{gathered}⟼Areaof△=s(s−a)(s−b)(s−c)</p><p></p><p>s = semi - perimeter</p><p></p><p>a = side</p><p></p><p>b = side</p><p></p><p>c = side</p><p></p><p>\begin{gathered}\\ \longmapsto\displaystyle \sf Area \: of \: \triangle = \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \\ \\ \\\end{gathered}⟼Areaof△=15(15−12)(15−12)(15−6)</p><p></p><p>\begin{gathered}\longmapsto\displaystyle \sf Area \: of \: \triangle = \sqrt{15 \times 3 \times 3 \times 9} \\ \\ \\\end{gathered}⟼Areaof△=15×3×3×9</p><p></p><p>\begin{gathered}\longmapsto\displaystyle \sf Area \: of \: \triangle = 3 \sqrt{135} \: cm ^{2} \\ \\\end{gathered}⟼Areaof△=3135cm2</p><p></p><p>\begin{gathered}\longmapsto\underline{ \boxed{\displaystyle \sf Area \: of \: \triangle = 9 \sqrt{15} \: cm ^{2} }} \\ \\\end{gathered}⟼Areaof△=915cm2</p><p></p><p>$