Which interaction of the atmosphere with the lithosphere benefits the biosphere?
2 answers:
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Based on scientific data, the stone is accelerated at approximately 10 .
<u>Given the following data:</u>
- Weight = 5 Newton
- Height = 44 meters.
Gravity can be defined as a force of attraction that controls the movement of the planets such as Earth around the Sun.
This ultimately implies that, gravity is a universal force of attraction that acts between all objects or matter that are having both mass and energy, and can occupy space.
<h3>Gravity on planet Earth</h3>Furthermore, the gravity near the Earth's surface generally makes it possible for all physical objects such as stones to possess weight and experience a free fall.
According to astronomical records, the acceleration due to gravity for an object experiencing a free fall near the Earth's surface is 9.8 .
In this scenario, the stone is accelerated at approximately 10 .
Read more on weight here: brainly.com/question/13833323
Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f =
D_f =
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m