(a) SG = SG_{(liquid)}x fraction submerged

\ fraction submerged = 0.58

(b) Rises (receives some upthrust from the water).

(a) P = r gh = 1025 x 10 x 100 = 1.025 MPa

F = PA = 1.025 x 10^{6}x 1.2 x 0.6 = 7.38 x 10^{5}N

(b) m = F/g = 7.38 x 10^{4}kg = 74 tonnes

NOTE: the diagram is missing from the textbook - it should look like this:

(a) container (ii); (b) container (ii); (c) container (ii); (d) all the same (pressure depends on the depth of the liquid, not the size of the base).

Firstly, calculate the time to strike ground ( s = ut + ½ at^{2}) and then calculate the horizontal velocity using v_{h }= Ö 2gh, then calculate the horizontal range of the liquid in that time using s_{h}= v_{h}t: eg for hole at a height of 30 cm (0.30 m):

0.30 = 0 + ½ x 10 x t^{2 }t = 0.245 s

v_{h}= Ö (2gh) = Ö (2 x 10 x 0.30) = 1.41 ms^{-1 }s_{h}= v_{h}t = 1.41 x 0.245 = 0.346 m

Full set of results:

Height (m)

t (s)

v

_{h}= Ö 2ghs

_{h}= v_{h}t0.30

0.245

1.41

0.346

0.20

0.200

2.00

0.400

0.10

0.141

2.45

0.345

The 30 cm and 10 cm holes have spouts that land in the same spot (0.346 m)

By Archimedes' Principle, the upthrust equals the weight of water displaced. The weight of the ice cubes = weight of the water displaced which is the volume of water below the water line. When the ice melts it will have a volume that fits into the space occupied by the ice that was underwater:

Hydrogen balloon:

Upthrust = V r = 400000 x 1.3/1000 x 10 = 5200 NHelium balloon:

Weight of balloon = 200 x 10 = 2000 N

Weight of gas = V r = 400000 x 0.09/1000 x 10 = 360 N

Result: Nett Force = 5200 - 2000 - 360 = 2840 N

Upthrust = V r = 400000 x 1.3/1000 x 10 = 5200 N

Weight of balloon = 200 x 10 = 2000 N

Weight of gas = V r = 400000 x 0.18/1000 x 10 = 720 N

Result: Nett Force = 5200 - 2000 - 720 = 2480 N

Conclusion: Hydrogen balloon lifts only 1.14 as much as the helium balloon.

Equal. The weight of the wood equals the upthrust which equals the weight of water displaced. The displaced water flows over the side of the bucket so the weight of the bucket is the same as before.

Nothing. The bucket and the water and the cork all fall at the same rate.

P = 1187 kgf /cm^{2}= 11870 N/cm^{2}= 1.187 x 10^{8}N/m^{2}(= Pa).

P = r gh

r = P/gh = 1.187 x 10^{8}/ (10 x 10916) = 1087 kg m^{-3 }

P = F/A = mg/A = r Vg/A = r hAg/A = r hg (the A cancels out)

P = 1.06 x 10^{3}x 1.83 x 10 = 19400 Pa (19.4 kPa).

P = 1/20 x 101.3 kPa = 5065 Pa

P = r gh

h = P/r g = 5065/(1000 x 10) = 0.5 m

SG = SG_{(fluid)}x fraction submerged

0.98 = SG x 2/3

SG = 1.47 g cm^{-3 }

Water level is still higher

The weight of water still has to equal the weight of mercury above the other level of mercury:

P_{water}= P_{mercury }r_{w}g h_{w}= r_{m}g h_{m}(cancel the g)

1 x h_{w}= h_{m}x 13.6

The water column is 13.6 times the height of the mercury.

Question 44

For both balloons: Upthrust = weight of air displaced;

Weight of air = mass of air x g

Mass of air = density of air x volume of balloon = D_{air}V_{ }Weight of air (upthrust) = D_{air}V x g

Weight of balloon and contents = (density of gas inside x volume x g) + (mass of balloon structure x g)

Weight (hydrogen balloon) = D_{H}V g + mg

Weight (helium balloon) = D_{He}V g + mg

Lift = Upthrust - weight

Lift (hydrogen) = D_{H}V g + mg - D_{air}V x g

Lift (helium) = D_{He}V g + mg - D_{air}V x g = 2D_{H}V g + mg - D_{air}V x g

Ratio: difference in lift/lift of helium

= ((-mg -D_{H}Vg + D_{air}Vg - (-mg -2D_{H}Vg + D_{air}Vg))/(-mg -2D_{H}Vg + D_{air}Vg)

= (-D_{H}Vg +2 D_{H}Vg)/(-mg - 2D_{H}Vg = D_{air}Vg)

= (D_{H}V)/(-m - 2D_{H}+ D_{air}) x 100%

8% = (0.089 x 100)/(-m - 0.178 +1.20)%

8 = 8.9/(-m + 1.022)

8m = 0.724

m = 0.09 kg

NOTE: answer in back of book says 2.13 kg but this is wrong. My mistake.

Question 45

Too hard. Need extra maths formulas. Ask your maths teacher.

Question 46

Too hard. Need extra maths formulas. Ask your maths teacher.