Answer:
the distance traveled by Charlotte in feet is 338.44 ft
Explanation:
Given;
speed of Charlotte, u = 66.5 mi/h
time of motion, t = 3.47 s
The distance traveled by Charlotte in feet is calculated as;

Therefore, the distance traveled by Charlotte in feet is 338.44 ft
Answer:
D. 2.8 × 10⁹ N
Explanation:
The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.
Fe= k Q₁Q₂/r²
Q₁= -0.0045 C
Q₂= -0.0025 C
r= 0.0060 m
k= 9.00 × 10 ⁹ Nm²/C²
Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²
=2.8 × 10⁹ N
Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

In case of perfectly inelastic collision v'1 and v'2 are same.
We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁
Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg
Mass of second mud ball is 6kg.