# 5 Physics Questions

1. An empty storage tank has a volume of 116 ft^{3}. What is the buoyant force exerted on it by the air?

2. The total mass of helium in a balloon is 108 kg. What is the volume of the balloon? Enter only the numeric portion of the answer, not the units.

3. A person with a weight of 288.9 lb stands on two feet with their weight equally distributed between the two shoes and each shoe has an area of 34.1 square inches. What is the pressure on **each shoe independently**?

4. Use Pascal’s principle. Given that F_{out} is 62 lb., the areas of the output and input cylinders are A_{out} 25 in^{2} and A_{in} 6 in^{2}, what is the input force?

5. A 1.6 kg ball is thrown upward at an initial velocity of 6 m/s. How high will the ball go?

1. The buoyant force exerted of a mass whose volume is V is by definition equal to the weight of the volume of the fluid displaced by that mass.

The density of air is $\rho_air = 1.2 kg/m^3$

$1 ft = 0.3 m$

$1ft^3 =0,027 m^3$

The buoyant force is $F = m*g =\rho_air*V*g = 1.2*0.027*116*9.81 =36.87 N$

2.The relation between the mass and the volume of a certain object is

Density = mass/volume.

In the case of Helium the density is

$\rho_He =0.179 kg/m^3$

Thus

Volume = mass/density $= 108/0.179 =603.35 m^3 =603350 L$

3.The pressure is by definition

Pressure = Force/Surface

Because there are two shoes the force on each shoe is half the weight.

$1 lb = 0.455 kg$

$1 inch = 2.54 cm$

$1 sq. inch = 6,4516 sq. cm$

The pressure is

$P =288.9*0.455*9.81/2/(34.1*6.4516) = 2.92 N/cm^2 = 2.92*10^{-4} N/m^2$

4. The Pascal principle says that the pressure exerted in an arbitrary point of a liquid is transmitted equally in all directions in that fluid.

The means the Pressure on the output cylinder is equal to the pressure in the input cylinder.

The pressure is by definition the force over the surface area.

Fout/Aout = Fin/Ain

Fin = Fout * Ain/Aout $= 62*6/25 = 14.88 lb = 14.88*0.455*9.81 =66.42 N$

5. The equation of uniform decelerated motion (with acceleration -g) for distances H is

$V^2 =V0^2 – 2*g*H$ (does not depend on the object mass)

where V is the final speed and V0 is the initial speed.

$0 = V0^2-2*g*H$

$H = v0^2/2/g = 6*6/2/9.81 =1.83 m$