Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass, 
Area of cross section, 
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
R = W
or

Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,

So, the drag coefficient is 1.01.
Answer:
Range, 
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
B) The amount of work done
Answer:
Explanation:
Of the 4 numbers given, the answer is 1 or A
If you take the absolute value of abs(1 - 1.04) you get 0.04.
(2 - 1.04) = 0.96
1.25 - 1.04 = .21
1.5 - 1.04 = 0.46
The last three are all larger than 0.04
Note: absolute value means the positive difference between 2 numbers (even though it is negative). If it is negative, absolute value makes it positive.
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds