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
D. Only B and C
Medium and High levels can be harmful. But low even levels aren't good either.
Answer:
<h2>1 st statement is correct </h2>
Explanation:
The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another. This means that a system always has the same amount of energy, unless it's added from the outside
<h2>hope you understand and got your answer by my explanation </h2><h2>thanks </h2>
<h2>please give brainliest plz follow </h2>
<span>For the orbitals that designate to be degenerate in a many-electron system, the following quantum numbers should be equal:
1- principle quantum number (n) : this describes the most probable distance of the electron from the nucleus
2- o</span><span>rbital angular momentum quantum number (l) : this determines the shape of the orbital </span>
Answer:
He will complete the race in total time of T = 10 s
Explanation:
Total distance moved by the sprinter in 2.14 s is given as
now the distance remaining to move
now he will move with uniform maximum speed for the remaining distance
so we will have
so the total time to complete the race is given as
