Given that,
Depth of seawater, h = 11,033 m
Density seawater, p (rho) = 1025 kg/m³
Gauge Pressure , P = ??
Since, we know that:
Pressure, P = pgh
Pressure = 1025 * 9.81 * 11033
Pressure = 1109395723.3 N/m²
or
Pressure = 1.1 x 10∧8 Pascal
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Answer:
22.15 N/m
Explanation:
As we know potential energy = m*g*h
Potential energy of spring = (1/2)kx^2
m*g*h = (1/2)kx^2
Substituting the given values, we get -
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
For safety reasons, this spring constant is increased by 13 % So the new spring constant is
k = 19600 * 1.13 = 22148 N/m = 22.15 N/m
At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m.
So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY.......
We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION:
v² = u² + 2as
0 = u² - 2gh
u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity)
So rearranging,
velocity = (velocity in y direction only) / sin 3°
= √(2gh)/sin 3°
= (√(2 x 9.8 x 0.33)) / sin 3°
= 49 m/s at 3° to the horizontal
