Three liquids that do not mix are poured into a cylindrical container with a diameter of 10.0 cm. The densities and volumes of the liquids are as follows. Liquid 1: 1 = 2.80 ✕ 103 kg/m3 and V1 = 2.00 ✕ 10−3 m3
Liquid 2: 2 = 1.00 ✕ 103 kg/m3 and V2 = 1.50 ✕ 10−3 m3
Liquid 3: 3 = 0.600 ✕ 103 kg/m3 and V3 = 1.00 ✕ 10−3 m3
Determine the pressure on the bottom of the container.

P = 9622.9 Pa = 9.62 KPa

Explanation:

First, we will calculate the mass of all three liquids:

m = ρV

where,

m = mass of liquid

ρ = density of liquid

V = Volume of liquid

FOR LIQUID 1:

m₁ = (2.8 x 10³ kg/m³)(2 x 10⁻³ m³) = 5.6 kg

m₂ = (1 x 10³ kg/m³)(1.5 x 10⁻³ m³) = 1.5 kg

m₃ = (0.6 x 10³ kg/m³)(1 x 10⁻³ m³) = 0.6 kg

The total mass will be:

m = m₁ + m₂+ m₃ = 5.6 kg + 1.5 kg + 0.6 kg

m = 7.7 kg

Hence, the weight of the liquids will be:

W = mg = (7.7 kg)(9.81 m/s²) = 75.54 N

Now, we calculate the base area:

A = πr² = π(0.05 m)²

A = 7.85 x 10⁻³ m²

Now the pressure will be given as:

P = 9622.9 Pa = 9.62 KPa

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.