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
Answer:
Kinetic energy of diver at 90% of the distance to the water is 9000 J
Explanation:
Let d is the distance between the position of the diver and surface of the pool.
Initially, the diver is at rest and only have potential energy which is equal to 10000 J.
As the diver dives towards the pool, its potential energy is converting into kinetic energy due to law of conservation of energy, as total energy of the system remains same.
Energy before diving = Energy during diving
(Potential Energy + Kinetic Energy) = (Kinetic Energy + Potential Energy)
When the diver reaches 90% of the distance to the water, its kinetic energy
is 90% to its initial potential energy, as its initial kinetic is zero,i.e.,
K.E. =
K.E. = 9000 J
Answer: The conductivity of water depends on the concentration of dissolved ions in solution. ... This is because the Sodium Chloride salt dissociates into ions. Hence sea water is about a million times more conductive than fresh water.
I think its A I hope this help thank you!!
(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
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