Answer:
1÷60 h
time equals distance upon speed
A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½ at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):
v^2 = u^2 + 2as
v^2 = 17.0^2 -2(9.81)(11.0)
v = √73.18 = 8.55m/s
now from equation (3):
v = u + at
8.55 = 17.0 – 9.81t
t = (8.55 – 17.0)/(-9.81) = 0.86s
The question can be solved using conservation of linear momentum.
= 0.06kg and 
= 0.03kg
Let the initial velocity of Marble A be , 
= 0.7m/s
Let the initial velocity of Marble B be, 
= 0m/s
Let the velocity of Marble A after collisiong , 
= -0,02m/s
Let the velocity of Marble B after collision be 
From the conservation of linear momentum equation. We get,
Substituting the values we get,
(0.06)(0.7) + 0 = (0.06)(-0.02) + (0.03)
we get, = 1.44m/s
Umm, I guess it really depends on what these numbers are for but what I can say is if you put this under physics then decimals are used very often and help express parts of numbers for this particular problem.And if it's directly between the two then it's 0.50 added to that smaller number,I hope that helps
Answer:
the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.
