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3 years ago

What will happen if more solute is added than can be dissolved .

Physics

1 answer:

zysi [14]3 years ago

3 0

The solute would then sink to the bottom and would not dissolve

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a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100

Anastaziya [24]

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

let's reduce the magnitudes to the SI system

r = 10 cm = 0.10 m

m = 10 g = 0.010 kg

A = 100 cm² = 0.01 m²

the definition of density is

ρ = m / V

the volume of a sphere

V = $\frac{4}{3} \ \pi r^{3}$

V = $\frac{4}{3}$ π 0.1³

V = 4.189 10⁻³ m³

let's calculate the density of the ball

ρ = $\frac{0.010}{4.189 \ 10^{-3} }$

ρ = 2.387 kg / m³

the tabulated density of water is

ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

B - W = 0

B = W

where B is the thrust that is given by Archimedes' principle

ρ_liquid g V_submerged = m g

V_submerged = m / ρ_liquid

we calculate

V _submerged = 0.10 9.8 / 997

V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

V = A h

h = V / A

let's calculate

h = 9.83 10⁻⁴ / 0.01

h = 0.0983 m

this is equal to h = 9.83 cm

8 0

2 years ago

The vector quantity that defines the distance and direction between two positions. It is a change in your position.

lilavasa [31]

I believe you're talking about displacement. It's a directional vector that depicts the movement of a point between two instances.

4 0

2 years ago

A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon

jek_recluse [69]

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

Στ = 0

W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

x₂ = $\frac{M_1g }{m_2 g} \ x_1$