Hi there! When a shadow is at its shortest is at noon when the sun is overhead.
THere is a standard relationship that gives this result where the capacity of the capacitor is used:

.
We know though that Q/c=V and thus we can use the relationship:
E=Q*V/2 where we have just substituted in. If we also take into account that Q=VC, we can also get that E=V^2*C/2.
We are given the charge and the potential, so the best expression to use is the middle one.
Substituting, we get that E=1/2*8*10^(-10)*20=8*10^(-9).
The answer is B
Answer:
Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from
The answer to the question is
The speed
of the electron when it is 10.0 cm from charge Q₁
= 7.53×10⁶ m/s
Explanation:
To solve the question we have
Q₁ = 3.45 nC = 3.45 × 10⁻⁹C
Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C
2·d = 50.0 cm
a = 10.0 cm
q = -1.6×10⁻¹⁹C
Also initial kinetic energy = 0 and
Initial electric potential energy = 
Final kinetic energy due to motion = 0.5·m·v²
Final electric potential energy = 
From the energy conservation principle we have

Solving for v gives

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg
gives v =7528188.32769 m/s or 7.53×10⁶ m/s
= 7.53×10⁶ m/s
Answer:
Terminal velocity is achieved, therefore, when the speed of a moving object is no longer increasing or decreasing; the object's acceleration (or deceleration) is zero.
Explanation:
hope it helps you