Draw a lift pump diagram and explain how it operates. Solve the following problems
A metal box of weigh 20 N rests on its 1 m by 0.6m side on floor. How
much is the pressure exerted by the metal box on the floor? Take ga
10 m/s?
A rectangular container of base 50 cm by 30 cm is filled with water to a
depth of 5 cm. How much is the pressure exerted at the base?
(Take pw= 1000 kg/m3 and g =10 m/s2)
How much is the pressure exerted at a point where a ball- pointed
pen of area 1mm? is pushed against the paper with a force of 24 N?
Suppose a scientist was able to construct a barometer with a liquid
being twice denser than mercury, then how high would the liquid raise
ard pressure?
m. if the
oil to a

1) The pressure exerted by the metal box on the floor is 33. Pa

2) The pressure exerted at the base is 250 Pa

3) The pressure exerted at the point where the pen pushes against the paper is 24 MPa

Explanation:

1) Pressure is given by the following relation

The force exerted by the weight of the metal box, F = 20 N

The dimensions of the side on which the metal box rests = 1 m by 0.6 m

∴ The area of the side on which the metal box rests, A = 1 m × 0.6 m = 0.6 m²

The pressure exerted by the metal box on the floor, 'P' is therefore given as follows;

The pressure exerted by the metal box on the floor, P = 33. Pa

2) The dimensions of the rectangular container are;

Base dimension = 50 cm by 30 cm = 0.5 m by 0.3 m

The depth to which the water is filled in the container = 5 cm = 0.05 m

The density of water, = 1,000 kg/m³

The volume of water in the container, V = Base area × Height = 0.5 m × 0.3 m × 0.05 m = 0.0075 m³

∴ V = 0.0075 m³

The mass of the water, m = V × = 0.0075 m³ × 1,000 kg/m³ = 7.5 kg

m = 7.5 kg

The weight of the water, W = m × g = 7.5 kg × 10 m/s² = 75 N

The base area, A = 0.5 m × 0.6 m = 0.3 m²

The pressure exerted at the base, P = W/A = 75 N/(0.3 m²) = 250 Pa

The pressure exerted at the base, P = 250 Pa

3) The area of the point of the ball-pointed pen pushing against the paper, A = 1 mm² = 0.000001 m²

The force with which the point on the pen pushes against the paper = 24 N

The pressure exerted at the point where the pen pushes against the paper, P = W/A

∴ P = 24 N/(0.000001 m²) = 24,000,000 Pa

The pressure exerted at the point where the pen pushes against the paper, P = 24,000,000 Pa = 24 MPa.

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

no sign of dia of fuel cell