D.
Sasha is informing participants about the study and making sure they wish to participate.
ANSWER:
300 J
STEP-BY-STEP EXPLANATION:
To know the work required, we must calculate the work in both cases, the difference would be the amount of work necessary for the speed to increase. The work done is the same as the amount of energy increase. The formula for kinetic energy is:

We calculate in each case:

We calculate the difference between the two to find out the work required:

The work required is 300 J
Falling from an airplane.
Answer:
Vd = 1.597 ×10⁻⁴ m/s
Explanation:
Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³
To find:
Drift Velocity Vd=?
Solution:
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)
Vd = 1.597 ×10⁻⁴ m/s
Answer:
a) Q1=Q2=480μC V1=240V V2=60V
b) Q1=96μC Q2=384μC V1=V2=48V
c) Q1=Q2=0C V1=V2=0V
Explanation:
Let C1 = 2μC and C2=8μC
For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:




For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:
So, 
Total charge is the same calculated for part (a), so:
Solving for Q2:
Q2 = 384μC Q1 = 96μC.
Therefore:
V1=V2=48V
For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C V1=V2=0V