D (Glucose +Oxygen --> Carbon Dioxide + Water + Energy)
Answer: The 6 kg rock sitting on a 3.2 m cliff.
Explanation:
The potential energy of an object of mass M that is at a height H above the ground us:
U = M*H*g
where g is the gravitational acceleration:
g = 9.8m/s^2
Then:
"An 8 kg rock sitting on a 2.2 m cliff"
M = 8kg
H = 2.2m
U = 8kg*2.2m*9.8 m/s^2 = 172.48 J
"a 6 kg rock sitting on a 3.2 m cliff"
M = 6kg
H = 3.2m
U = 6kg*3.2m*9.8m/s^2 = 188.16 J
You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.
F = qE + qV × B
where force F, electric field E, velocity V, and magnetic field B are vectors and the × operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = -V × B
which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get
|V| = 8.8e3 / 3.7e-3
= 2.4e6 m/sec
Acceleration a = V²/r, where r is the radius of curvature.
a = F/m, where m is the mass of an electron,
so qVB/m = V²/r.
Solving for r yields
r = mV/qB
= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)
= 3.65e-3 m
Electricity flows from positive to negative
