Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
Answer:
132.9 cm
Explanation:
Data provided in the question:
Radius of the basll = 76 cm = 0.76 m
Side of the box = 200 cm = 2 m
Density of the ball and cube are equal
let the density be 'D'
Now,
Mass of ball, M = Volume × Density
=
× D
=
× D
= 1.838D
Mass of cube, m = L³ × D
= 2³ × D
= 8D
Thus,
center of mass, y = [ My₁ + my₂ ] ÷ [M + m]
here,
y₁ = center of mass of ball with respect to floor
as the center mass of sphere lies at the center of the sphere
= Length of cube + radius of sphere
= 2 + 0.76
= 2.76 m
y₂ = Center of mass of cube =
= 1 m
Thus,
y = [ ( 1.838D × 2.76 ) + (8D × 1 ) ] ÷ [1.838D + 8D]
= 13.07288D ÷ 9.838D
= 1.329 m
or
= 132.9 cm
Answer:<span>Humid air is lighter, so it has lower pressure.
The reason is the molecules of water are H2O, whose molar mass is 18 g/mol.
These molecules displaces molecules of N2 and O2, whose molar masses are:
N2: 2*14g/mol = 28 g/mol, and
O2: 2*16g/mol = 32 g/mol.
Then molecules of 28g/mol and 32 g/mol are being replaced with molecules of 18g/mol, leading to a lower weight of the same volume of air, which results in lower pressure.
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