Please answer number 3. Physics 1441: Homework 25 A Styrofoam board 10.0 cm thic
ID: 2260743 • Letter: P
Question
Please answer number 3.
Physics 1441: Homework 25 A Styrofoam board 10.0 cm thick has a density of 300 kg/m3. The board floats in water with its top at the water's surface when a swimmer is resting on it. If the area of the board is 1.00m2, what is the mass of the swimmer? What is the pressure on the bottom of the ocean floor at a depth of 5,000m? PIston1 (left) has a diameter of 0.25cm and piston 2 has a diameter of 1.5cm. In the absence of friction, determine the force F necessary to support a 500N weight.Explanation / Answer
1. Let mass of swimmer be Ms
Mass of board= Mb= 300kg/m3 X 0.1m X 1m2= 30kg.
Weight of board= 30kg X9.81m/s2=294.3N
Buoyant force on the board= 1000kg/m3 X 9.81m/s2 X 0.1m X 1m2=981N
Weight of swimmer= 981-294.3= 686.7N
Mass of swimmer= 686.7/9.81= 70kg.
2. Pressure at ocean bottom= P= Po + (Density X g X H)
Po= atmospheric pressure= 100000 Pa
Density=1030 kg/m3
H=5000m, g=9.81m/s2
Substituting all values we get P=50621500Pa.
3. As the hieght difference between the two pistons is not mentioned we neglect the height diff.
Therefore, for equlibrium Pressure applied by piston 1 on the fluid = Pressure applied by piston2
P= F/A
P1=F1/A1, P2=F2/A2
P1=P2 implies F1/A1=F2/A2.
A1=(pi*d*d)/4=0.04909 cm2. (d=0.25cm)
A2=(pi*D*D)/4=1.76714cm2. (D=1.5cm)
F2=500N.
Therefore, F1=(F2*A1)/A2=13.889N.
Now taking moment about the pivot of the handle=0
F1*2= F*12
F=F1/6
F=2.315N