What do the four main layers of the atmosphere have in common
1 answer:
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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 =
V = π 0.1³
V = 4.189 10⁻³ m³
let's calculate the density of the ball
ρ =
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
Hello!
Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:
← The circumference of the orbit
speed = orbital speed, we will solve for this later
time = period
Therefore:
Where 'r' is the orbital radius of the satellite.
First, let's solve for 'v' assuming a uniform orbit using the equation:
G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)
m = mass of the earth (5.98 × 10²⁴ kg)
r = radius of orbit (1.276 × 10⁷ m)
Plug in the givens:
Now, we can solve for the period: