1. ans: 3.4 x105 N;
Area of pool is p m2.
Total weight of water is p*1000*9.8 N.
Total weight of air above pool is 10100*p N.
2. ans: 21.5 grams/cm3;
Density = mass /volume.
Don't forget to use consistent units!
4. ans: 600 N;
Pressure is constant, so force is proportional to area (F=PA).
5. ans: 83;
Every meter of water depth adds 9800
N/m2 of pressure (1000*9.8).
6. ans: 0.968 x105 Pa;
The pressure of mercury is 0.726*1*1*density*g N/m2.
7. ans: 140,000 N;
0.7 ATM (= pressure difference) times the area.
8. ans: 1960 Pa;
Mass*gravity/surface area.
9. ans: 563 gm;
First determine the volume of aluminum, =mass/density
(gives cm3 here), and then the mass of oil with the
same volume = oil.density*volume,
and then subtract the oil mass from the aluminum
mass. The buoyancy force always = - weight of displaced fluid.
10. ans: 0.258 N;
The net force is the weight of the displaced water less the weight of
the ball, = [(water density)*(ball volume)
-ball mass grams]*0.001*9.8 N. the .001 converts from grams to kg.
11. ans: 2.2 cm;
The same mass of water occupies 0.78 times the volume of the wood,
so the height under water will be 7.8 cm.
12. ans: 0.015 m;
The volume of water with mass equal to 60 kg is 60kg/1000kg/cm3,
and the depth*surface area also = this volume.
13. ans: 1580 kg/m3;
The mass of water displaced must equal the difference of measured masses,
5.7 kg. Then the volume displaced must be (5.7/1000) m3.