Ek = 1/2 mv^2
9 × 10^4 = 1/2 × 800 × v^2
9 × 10^4/400 = 400 v^2 / 400
9 × 10^4/400 = v^2
√225 = v
15 ms⁻¹ = v
That's the only way I know how to work it out
I think in this case velocity and speed would be considered the same because me
s = d/t and v=d/t
one is distance travelled and the other is displacement of a body
Answer:

Explanation:
The charge on one object, 
The distance between the charges, r = 0.22 m
The force between the charges, F = 4,550 N
Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

So, the charge on the other sphere is
.
K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2
So, option B is your answer!
Hope this helps!
She could text/email,report it to them personally,post up flyers,tell one person to spread it around
Answer:
After 9 seconds the object reaches ground.
Step-by-step explanation:
We equation of motion given as h = -16t²+128t+144,
We need to find in how many seconds will the object hit the ground,
That is we need to find time when h = 0
0 = -16t²+128t+144
16t²-128t-144= 0

Negative time is not possible, hence after 9 seconds the object reaches ground.