if a depth of liquid of 1.0meter causes pressure of 7 kpa,what is the specific gravity of the liquid
"p=\\rho gh; \\rho=\\frac{p}{gh}=\\frac{7\\times10^3}{9.8\\times 1}=714.29\\frac{kg}{m^3}\\\\or\\ 0.71429\\frac{g}{cm^3}.\\\\Answer:the\\ specific\\ gravity\\ of\\ the\\ liquid\\\\ is\\;714.29\\frac{kg}{m^3}\\;or\\ 0.71429\\frac{g}{cm^3}."
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