Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.Displacement<span> is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
</span>To calculate displacement<span>, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your </span>displacement<span> is 0. In physics, </span>displacement<span> is represented by Δs.
For me to solve this I would need to know the time, but I can give you a handy displacement calculator I used that helped me.
https://www.easycalculation.com/physics/classical-physics/constant-acc-displacement.php
Hope I helped.
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Answer:
F=X.F=mxq. 3. 1 N/kg=0.5kg g=9.80
The only answer that can justify being a hypothesis is C.
The evidence that the universe is expanding comes with something called the red shift<span> of light. Light travels to Earth from other galaxies. As the light from that galaxy gets closer to Earth, the distance between Earth and the galaxy increases, which causes the wavelength of that light to get longer.</span>
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>
