<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>
Answer:
Explanation:
Let pressure at surface of earth be P Pa.
pressure at height of 8.1 km in air can be calculated as follows .
pressure due to column of air of 8.1 km height
= h d g , h is height , d is density of air and g is acceleration due to gravity
= 8.1 x 1000 x .87 x 9.8 = 6.9 x 10⁴ Pa .
pressure at the height of 8.1 km
= P - 6.9 x 10⁴ Pa
Pressure due to column of 16 m in the sea
= h d g
16 x 1000 x 9.8
= 15.68 x 10⁴ Pa .
Pressure at depth of 16m
= P + 15.68 x 10⁴
pressure difference between points at height of 8.1 km and pressure at point 16 m deep
= P + 15.68 x 10⁴ - P + 6.9 x 10⁴ Pa
= 22.58 x 10⁴ Pa .
You need to convert minutes to hours so
