pg 55-56
1. A [gas supply on right pressing more ==> gas pressure has to be higher]
2. C [pressure of liquid = hpg]
3. D
Pressure on left = Pressure on right [pressure in hydraulic fluid same throughout]
F ÷ A = F ÷ A
10 N ÷ 10 cm² = (1500 x 10) ÷ A
A = 15000 cm²
4. A
Let weight of small cylinder be W and area of small cylinder be A
Therefore weight of big cylinder = 6W and area of big cylinder = 4A
Pressure of big cylinder = 6W ÷ 4A = 18 Pa
therefore 1.5 (W ÷ A) = 18 Pa
(W ÷ A) = 18 ÷ 1.5 = 12 Pa
Pressure of small cylinder = W ÷ A = 12 Pa
Section B
1b.
i. Weight on ground by each foot = 20000 N ÷ 4 = 5000 N
[since 4 feet, so weight is equally distributed among the 4]
Pressure = F ÷ A = 5000 N ÷ 200 cm² = 5000 N ÷ 0.02 m² = 250000 Pa
Alternatively, since pressure on one foot = pressure on all feet = 20000 N ÷ (4 x 200) cm² = 250000 Pa
ii. A = F ÷ Pressure = 600 N ÷ 250000 Pa = 0.0024 m²
2a.
i. Pressure at surface = atmospheric pressure = 101000 Pa
ii Pressure at 5m below = liquid pressure + Patm = hpg + 101000 = 5 x 1025 x 5 + 101000 = 152250 Pa
2b. 152250 - 101000 = 51250 Pa
3.
a. Pressure on small piston = F ÷ A = 120 N ÷ 20 cm² = 120 N ÷ 0.0020 m² = 60000 Pa
b. Pressure on large piston = pressure on small piston = 60000 Pa
c. Weight = Pressure x A = 60000 Pa x (300 cm² ÷ 10000) m² = 1800 N
4
a. Atmospheric pressure is equivalent to the pressure exerted by a mercury column of vertical height 76.0 cm
b. Pressure = hpg = 0.760 m x 13600 x 10 = 103360 Pa = 103000 Pa (3sf)
5a.
Pressure at point O = 0 Pa
Pressure at point P = hpg = (3h ÷ 4 ÷ 100) m x 13600 x 10 = 1020h Pa
Pressure at point Q = hpg = (h ÷ 100) m x 13600 x 10 = 1360h Pa
5b.
Shorter length of barometer as mercury higher density than water
Easier to observe as mercury is opaque, water is colourless and transparent [can add colouring though]
More accurate reading as mercury does not stick to glass, whereas water does
More accurate reading as mercury has a lower evaporation rate than water
6a.
Mercury in left arm will go down, in right arm will go up.
Air pressure in balloon higher than atmospheric pressure, pushes the mercury in left arm down.
[think about releasing a balloon filled with air, the balloon will sort of force the air out right? This sort of exert higher pressure. Also, the balloon will shrink after it is attached to the manometer.
*the reason why the balloon grows bigger is to reduce the pressure of air inside to be the same as atmospheric pressure, but given a chance, the balloon will want to go back to its original size]
6b
mercury in left arm will go up, in right arm will go down, until both same level
6c
use liquid of lower density (example water, alcohol, etc)
7a
For Figure 7.4, pressure in mmHg = Patm +h = 760 +100 = 860 mmHg
For Figure 7.5, pressure in mmHg = Patm - h = 760 - 50 = 710 mmHg
7b
compare densities of immiscible liquids
8a
mass of X = density x volume = 2.00 x (2 x 10 ) = 40 g
weight of X = mg = 0.040 kg x 10 = 0.4 N
8b
Pressure = F ÷ A = 0.4 N ÷ 2 cm² = 0.4 N ÷ 0.0002 m² = 2000 Pa
8c
mass of Y = density x volume = (5.00 x A x h) g
weight of Y = (5.00 x A x h ÷ 1000) x 10 = 0.05 Ah N
Pressure of Y = Weight of Y ÷ A
2000 = 0.05 Ah ÷ (A ÷10000) = 500 h
h of Y = 4 cm
mass of Z = density x volume = (10.00 x A x h) g
weight of Z = (10.00 x A x h ÷ 1000) x 10 = 0.10 Ah N
Pressure of Z = Weight of Z ÷ A
2000 = 0.10 Ah ÷ (A ÷10000) = 1000 h
h of Z = 2 cm
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