It is D, you can not replace minerals. This makes them valuable.
Force=mass x acceleration
f= 0.5 x40
f=20N
Answer:
112 m/s², 79.1°
Explanation:
In the x direction, given:
x₀ = 0 m
x = 19,500 cos 32.0° m
v₀ = 1810 cos 20.0° m/s
t = 9.20 s
Find: a
x = x₀ + v₀ t + ½ at²
19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²
a = 21.01 m/s²
In the y direction, given:
y₀ = 0 m
y = 19,500 sin 32.0° m
v₀ = 1810 sin 20.0° m/s
t = 9.20 s
Find: a
y = y₀ + v₀ t + ½ at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²
a = 109.6 m/s²
The magnitude of the acceleration is:
a² = ax² + ay²
a² = (21.01)² + (109.6)²
a = 112 m/s²
And the direction is:
θ = atan(ay / ax)
θ = atan(109.6 / 21.01)
θ = 79.1°
It considered as Zero Gage pressure.
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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>
