Answer:
The third drop is 0.26m
Explanation:
The drop 1 impacts at time T is given by:
T=sqrt(2h/g)
T= sqrt[(2×2.4)/9.8]
T= sqrt(4.8/9.8)
T= sqrt(0.4898)
T= 0.70seconds
4th drops starts at dT=0.70/3= 0.23seconds
The interval between the drops is 0.23seconds
Third drop will fall at t= 0.23
h=1/2gt^2
h= 1/2×9.81×(0.23)^2
h= 0.26m
I think it’s, its direction would reinforce the original movement
Answer:
(a) A. Uniform line of charge and B. Uniformly charged sphere
(b) To three digits of precision:
λ = 1.50 * 10^-10 C/m
p = 2.81 * 10^-4 C/m^3
Explanation:
First, let's express the movement of Car A and B in terms of their position over time (relative to car B)
For car A: y=20x-200 Car A moves 20 meters every second x, and starts 200 meters behind car B
For Car B: y= 15x Car B moves 15 meters every second and starts at our basis point
Set the two equations equal to one another to find the time x at which they meet:
20x - 200 = 15x
200 = 5x
x= 40
At time x=40 seconds, the cars meet. How far will Car A have traveled at this time?
Car A moves 20 meters every second:
20 x 40 = 800 meters
