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:
Explanation:
The cross product of two vectors is given by
Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.
Here, K x i = - j
As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along axis.
The direction of cross product of two vectors is given by the right hand palm rule.
So, k x i = j
j x i = - k
- j x k = - i
i x i = 0
Answer:
221.17 kJ
Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.
N = 97.6 ÷ 18
Q=molar heat *moles
Q = (40.79) * (97.6 ÷ 18)
This is approximately 221.17 kJ
Explanation:
Below is an attachment containing the solution.
Answer:
Explanation:
For resistance of a wire , the formula is as follows .
R = ρ L/S
where ρ is specific resistance , L is length and S is cross sectional area of wire .
for first wire resistance
R₁ = ρ 3L/3a = ρ L/a
for second wire , resistance
R₂ = ρ 3L/6a
= .5 ρ L/a
For 3 rd wire resistance
R₃ = ρ 6L/3a
= 2ρ L/a
For fourth wire , resistance
R₄ = ρ 6L/6a
= ρ L/a
So the smallest resistance is of second wire .
Its resistance is .5 ρ L/a
