Answer:
The order of magnitude of the distance from the sun to Earth is 10⁸ km.
Explanation:
The order of magnitude of the distance from the sun to Earth can be calculated as follows:
Where:
c: is the speed of light = 3x10⁸ m/s
t: is the time = 8 min
Hence, the distance is:
Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.
I hope it helps you!
It's the angle made by the incident ray when it's perpendicular to the surface. (Perpendicular lines are the lines that form a graph or like a 90-degree angle)
The area under the velocity time graph is 125 m and the meaning of the area is displacement.
<h3>
What is area under velocity - time graph?</h3>
The area under a velocity time graph represents the displacement of the object.
total area of the graph = A1 + A2
total area of the graph = ¹/₂ (base₁)(height₁) + ¹/₂ (base₂)(height₂)
total area of the graph = ¹/₂(4)(40) + ¹/₂(3)(30)
total area of the graph = 125 m
Thus, the area under the velocity time graph is 125 m and the meaning of the area is displacement.
Learn more about velocity time graph here: brainly.com/question/4710544
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A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² -
r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄
B) E = kQ/d²
since the distribution is symmetric spherically
C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 -
r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.
D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a
min).
<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>