Physics help! Fluids?

1. The density of gold is 19 300 kg/m3. What is the mass of a gold sphere whose diameter is 3.15 cm?2. A child wants to pump up a bicycle tires so tha

1. The density of gold is 19 300 kg/m3. What is the mass of a gold sphere whose diameter is 3.15 cm?

2. A child wants to pump up a bicycle tires so that its pressure is 1.25 x 10^5 Pa above that of atmospheric pressure. If the child uses the pump with a circular piston 0.0285 m in diameter what force must the child exert?

3. A pirates treasure chest lies 8.33 m below the surface of the hidden lagoon
a) What is the pressure acting on the treasure chest?
b) What is the magnitude of the force that acts on the rectangular treasure chest top that is 0.651 m x 0.393 m?

or:1. The density of gold is 19 300 kg/m3. What is the mass of a gold sphere whose diameter is 3.15 cm?2. A child wants to pump up a bicycle tires so that its pressure is 1.25 x 10^5 Pa above that of atmospheric pressure. If the child uses the pump with a circular piston 0.0285 m in diameter what force must the child exert?3. A pirates treasure chest lies 8.33 m below the surface of the hidden lagoon a) What is the pressure acting on the treasure chest?b) What is the magnitude of the force that acts on the rectangular treasure chest top that is 0.651 m x 0.393 m?

or:1.volume of a sphere is 4/3 *pi *radius^3V = 4/3 * pi * (3.15 * 10^ -2 m /2)^3m = \u03c1 * V= 19 300 kg/m^3 * 4/3 * pi * (3.15 * 10^ -2 m /2)^3= 0.316 kg (3 sig.fig.)2Technically we should be considering the work being done by the child dW = dV * dP or dW = dF * dsHowever, this doesn't seem to be the purpose of the questionAnyway, apart from the question making absolutely no sense whatever to me, I can still apply a formuladF = dP * dAdF= 1.25 * 10^5 Pa * pi * (0.0285 m /2)^2 = 79.7 N (3 sig.fig)3.The pressure exerted by a static fluid is dependent on gravity, density of fluid and depthP = g * \u03c1 * hDensity of lagoon water is a bit tricky, they're usually salty, but not as much as sea waterSea water can be as dense as 1029 kg/m^3Fresh water up to 999.97 kg/m^3Let's assume the lagoon water is two thirds as saturated at 1020 kg/m^3P = 9.8 m/s^2 * 1020 kg/m^3 * 8.33 m = 83 300 pa (3 sig.fig.)b) F = P * A = 83266.68 pa * 0.651 m * 0.393 m= 21 300 N

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