More expensive & probably less complicated
There are 2.1077x10^24 atoms of sulfur in 3.50 mols of sulfur.
Using the given formula with v0=56 ft/s and h=40 ft
h = -16t2 + v0t
40 = -16t2 + 56t
16t2 - 56t + 40 = 0
Solving the quadratic equation:
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32
We have two possible solutions
t1 = (56+24)/32 = 2.5
t2 = (56-24)/32 = 1
So initially the ball reach a height of 40 ft in 1 second.
Answer:
0.707m
Explanation:
from formula of range i.e R=Usin2Q/g
Answer: D. 0.29 m
Explanation:
We will use the following equations to describe the leap of the cat:
(1)
(2)
Where:
is the height of the cat
is the cat's initial velocity

is the acceleration due gravity
is the time
is the y-component of the velocity
Now the cat will have its maximum height
when
. So equation (2) is rewritten as:
(3)
Finding
:
(4)
(5)
(6)
Substituting (6) in (1):
(7)
Finally:
(8)