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Salsk061 [2.6K]
4 years ago
6

You have two points in a soil. Point A is at 75 cm; point B is at 25 cm above the reference. The capillary potential energy at p

oint A is -100 cm. The capillary potential energy at point B is -200 cm. What is the total potential energy at point A?
Physics
1 answer:
user100 [1]4 years ago
8 0

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

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Dimension of<br>Upthrust<br>​
alina1380 [7]

Explanation:

upthrust is the same as Force

And the dimension of Force is MLT-2

The -2 is going to be in raise to power form....

Hope it helps....

8 0
3 years ago
1) La corriente de un inductor de 90.0 mH cambia con el transcurso del tiempo de la forma I = 1.00 t 2 – 6.00 t (en unidades del
Gala2k [10]

Answer:

a)   E_{L} = -360 V , b)      t = 3 s

Explanation:

The electromotive force in an inductor is

           E_{L} = - L dI/ dt

in the exercise they give us the relation of i (t)

            i (t) = 1.00 t² -6.00t

we carry out the derivative and substitute

          E_{L} = - L (2.00 2t - 6.00 1)

a) the electromotive force at t = 1.00 s

          E_{L} = - 90.0 (4.00 1 - 6.00)

         E_{L} = -360 V

b) for t = 4 s

         E_{L}= - 90 (2 4 2 - 6 4)

         E_{L} = - 720 V

c) for the induced electromotive force to zero, the amount between paracentesis must be zero

           (2.00 t2 - 6.00t) = 0

            t (2.0 t-6.00) = 0

the solutions of this equation are

            t = 0

            2 t -6 = 0

            t = 3 s

to have a different solution the trivial (all zero) we must total t = 3 s

4 0
3 years ago
Butter is an example for<br>liquid in solid or solid in liquid ​
ioda

Answer:

Liquid in solid.

Explanation:

Butter is colloidal in nature belonging to the category of gels. This means that the dispersed phase is liquid and the dispersing medium is solid. In short, butter is an example for liquid in solid.

8 0
3 years ago
An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2
Gekata [30.6K]

Force on a moving charge is given by formula

\vec F = q(\vec v \times \vec B)

here we know that this force will be maximum when velocity is perpendicular to magnetic field

\vec F = qvB

here we know that

v = 2.0 \times 10^5 m/s

q = 1.6 \times 10^{-19} C

B = 8.5 \times 10^2 T

now we have

F = (1.6 \times 10^{-19})(2 \times 10^5)(8.5 \times 10^2)

F = 2.72 \times 10^{-11} N

7 0
4 years ago
An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

4 0
3 years ago
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