The boiling point of water, or any liquid, varies according to the surrounding atmospheric pressure. A liquid boils, or begins turning to vapor, when its internal vapor pressure equals the atmospheric pressure.
<span>The diver is heading downwards at 12 m/s
Ignoring air resistance, the formula for the distance under constant acceleration is
d = VT - 0.5AT^2
where
V = initial velocity
T = time
A = acceleration (9.8 m/s^2 on Earth)
In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m)
So let's substitute the known values and solve for T
d = VT - 0.5AT^2
-7 = 2.5T - 0.5*9.8T^2
-7 = 2.5T - 4.9T^2
0 = 2.5T - 4.9T^2 + 7
We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164.
Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So
V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s
V = 2.5 m/s - 14.47706141 m/s
V = -11.97706141 m/s
So the diver is going down at a velocity of 11.98 m/s
Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point.
V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s)
V = 2.5 m/s - (-9.477061409 m/s)
V = 2.5 m/s + 9.477061409 m/s
V = 11.97706141 m/s
And you get the exact same velocity, except it's the opposite sign.
In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
This situation describes the Hooke's Law which states that "When an elastic object - such as a spring - is stretched, the increased length is called its extension. The extension of an elastic object is directly proportional to the force applied to it". The formula is <span>F = k × e , F for the force, k for spring constant expressed in N/m, e for extension in m. This equation works for as long the spring is not stretch too much because once it exceeded its limit, the spring will not return to its original length the moment the load is removed.</span>
The rectilinear motion is given as:
s = s0 + v0 t + (1/2) a t^2
We can see that the variables stand for:
s = final distance
s0 = initial distance
v0 = initial velocity
t = time
a = acceleration
<span>We can see that all these variable are always constant
EXCEPT acceleration. The rectilinear motion can only be applied if and only if
acceleration is CONSTANT.</span>
