<span>Depth = 5.0 Ă— 10^2 m
Density of sea water = 1.025 x 10^3
Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa
Since the normal pressure is retained in the hull, no need to bother about Po
Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
So it is 49.56 times larger.</span>
Answer:
The potential energy at point A is 17.1675 J
Explanation:
The capillary potential is the work expended to bring up a unit mass of liquid to a point in a capillary region from a level liquid surface. It is the capillary potential that facilitates the movement of moisture within soil capillaries
In meteorology it is used to describe the level of saturated soil above the water table
Potential energy is the energy inherent in a body by virtue of its position, therefore the potentials of both point A and B are
Point A, elevation = 75 cm capillary potential = -100 cm
Point B, elevation = 25 cm capillary potential = -200 cm
The total potential energy at point A is
Elevation above reference - capillary potential =75-(-100) = 175 cm
which gives per unit mass
PE = m × g × h = 1 kg × 9.81 m/s ² × 1.75 m = 17.1675 kg·m²/s² = 17.1675 J
2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.
Answer:

Explanation:
<u>Given Data:</u>
Mass = m = 4 kg
Acceleration due to gravity = g = 9.8 m/s²
Height = h = 1 m
<u>Required:</u>
Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = mgh
<u>Solution:</u>
P.E. = (4)(9.8)(1)
P.E. = 39.2 Joules
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
A particle smaller than an atom or a <span>cluster of such particles </span>