Express your answer in pascals to three significant figures. h1=13.5 cm h2=6.00 cm mercury of density= 1.36×10 4 kg/m 3 Solve the following differential equation: y” + 4y = sin^3(x) Find answer of form: y(x)=________+C1________+C2________ .
Answer:- Well, let’s start by finding the pressure due to the “extra” height of the mercury. p = 1.36e4 kg/mÂł · (0.105m – 0.05m) · 9.8m/s² = 7330 N/m² = 7330 Pa The pressure at B is clearly p_b = p_atmos = p_gas + 7330 Pa The pressure at A is p_a = p_gas = p_atmos – 7330 Pa c) 1 atm = 101 325 Pa Then p_gas = 101325 Pa – 7330 Pa = 93 995 Pa
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