For a stock to be in equilibrium, that is, for there to be no long-term pressure for its price to depart

For a stock to be in equilibrium, that is, for there to be no long-term pressure for its price to depart from its current level, then a. the expected future return must be less than the most recent past realized return. b. the past realized return must be equal to the expected return during the same period. c. the expected future returns must be equal to the required return. d. the required return must equal the realized return in all periods. e. the expected return must be equal to both the required future return and the past realized return.

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  1. Pressure is defined as a measure of the force applied over a unit area. Pressure is often expressed in units of Pascals (Pa), newtons per square meter (N/m2 or kg/m·s2), or pounds per square inch

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  2. c.the expected future returns must be equal to the required return.

    Explanation:

    When the stock is at equilibrium than the intrinsic value of the stock is equivalent to the market price of the stock that depicts that the expected returns which held in the future should be equivalent to the required return

    Therefore the option c is correct

    And, the other options that are mentioned in the question are incorrect

  3. Pressure drop across the contraction section = 133 kPa

    The pressure difference due to frictional losses is = 39.7 kPa

    The pressure difference due to kinetic energy changes = 93 kPa

    Explanation:

    If [tex]V = \frac{Q}{A}[/tex]      -----------equation (1)

    where ;

    V  = velocity

    Q = flow rate

    A = area of cross-section

    As we know that Area (A)  = [tex](\frac{\pi }{4}D^2)[/tex]

    substituting  [tex](\frac{\pi }{4}D^2)[/tex] for A in equation (1); we have:

    [tex]V= \frac{Q}{\frac{\pi }{4}D^2 }[/tex]

    [tex]V= \frac{4Q}{\pi D^2 }[/tex] ----------------- equation (2)

    Now, having gotten that; lets find out the corresponding velocity [tex](V_1)[/tex] of the water at point (1) of the pipe and velocity [tex](V_2)[/tex] of the water at point 2 using the derived formula.

    For velocity [tex](V_1)[/tex] :

    [tex]V_1= \frac{4Q_1}{\pi D_1^2 }[/tex]

    From, the question; we are given that:

    water flow rate at point 1  [tex](Q_1)[/tex] = [tex]0.040m^3/s[/tex]

    Diameter of the pipe at point 1 [tex](D_1)[/tex] = 0.12 m

    ∴ [tex]V_1= \frac{4(0.04m^3/s)}{\pi (0.12m/s)^2 }[/tex]

    [tex]V_1=3.5367 m/s[/tex]

    For velocity [tex](V_2)[/tex]:

    [tex]V_2= \frac{4Q_2}{\pi D_2^2 }[/tex]

    water flow rate at point [tex](Q_2)[/tex] = [tex]0.040m^3/s[/tex]

    Diameter of the pipe at point 2 [tex](D_2)[/tex] = 0.06 m

    [tex]V_2= \frac{4(0.04m^3/s)}{\pi (0.06m/s)^2 }[/tex]

    [tex]V_2=14.1471 m/s[/tex]

    Similarly, since we have found out our veocity; lets find the proportion of the area used in both points. So proportion of   [tex](\frac{A_2}{A_1})[/tex]   can be find by replacing [tex](\frac{\pi }{4}D_2^2)[/tex] for [tex]A_2[/tex] and [tex](\frac{\pi }{4}D_1^2)[/tex] for [tex]A_1[/tex].

    So:   [tex]\frac{A_2}{A_1} = \frac{\frac{\pi }{4}D^2_2 }{\frac{\pi }{4}D^2_1 }[/tex]

    [tex]\frac{A_2}{A_1} = \frac{D_2^2}{D_1^2}[/tex]

    [tex]\frac{A_2}{A_1} = \frac{(0.06m)^2}{(0.12m)^2}[/tex]

    = 0.25

    However, let's proceed to  the phase where we determine the pressure drop across the contraction  Δp by using the expression.

    Δp = [tex][\frac{p_{water}}{2} (V_2^2-V_1^2)+ \frac{p_{water}}{2} K_LV_2^2][/tex]

    where;

    [tex]K_L[/tex] = standard frictional loss coefficient for a sudden contraction which is 0.4

    [tex]p_{water[/tex] = density of the water = 999 kg/m³

    [tex]\frac{p_{water}}{2} K_LV_2^2[/tex] = pressure difference due to frictional losses.

    [tex]\frac{p_{water}}{2} (V_2^2-V_1^2)[/tex] = pressure difference due to the kinetic energy

    So; we are calculating three terms here.

    a)  the pressure drop across the contraction = Δp

    b)  pressure difference due to frictional losses. = [tex]\frac{p_{water}}{2} K_LV_2^2[/tex]

    c)  pressure difference due to the kinetic energy =   [tex]\frac{p_{water}}{2} (V_2^2-V_1^2)[/tex]

    a)    Δp  =  [tex][\frac{p_{water}}{2} (V_2^2-V_1^2)+ \frac{p_{water}}{2} K_LV_2^2][/tex]

          Δp  =  [tex][\frac{999kg/m^3}{2} ((14.1m/s)^2-(3.53m/s)^2)+ \frac{999kg/m^3}{2} (0.4)(14.1m/s)^2][/tex]

          Δp  =  [(93081.375) + (39722.238)]

          Δp  =  (93 kPa) + (39.7 kPa)

          Δp  =  132.7 kPa

          Δp  ≅  133 kPa

    ∴ the pressure drop across the contraction  Δp = 133 kPa

      the pressure difference due to frictional losses [tex]\frac{p_{water}}{2} K_LV_2^2[/tex]  = 39.7 kPa

      the pressure difference due to the kinetic energy [tex]\frac{p_{water}}{2} (V_2^2-V_1^2)[/tex]  = 93 kpa

    I hope that helps a lot!

  4. 1) Hiatal hernia is the disorder of this 26 year old business executive. Hiatal hernia occurs when upper part of stomach pushes through an opening in the diaphragm, and up into the chest. THis opening is called a esophageal hiatus or diaphragmatic hiatus. It is basically a protrusion of the upper part of stomach into the chest through a tear on the chest through a tear or weakness in the diaphragm. The patient presents with :

    Chest painSevere hurtburnAbdominal painFrequent burpingDifficulty in swallowingThroat sorenessBelchingNausea

    2) Adequate lower esophageal pressure at the lower esophageal sphincter normally prevent gastric reflux into the esophagus when lying down or bending over.

    3) As M3 receptors (parasympathetic) are distributed at the lower esophageal sphincter, cholinergic agonist activates these receptors and increases contraction thus rsulting in a decrease in contraction and preventing Gastroesophageal reflux. Similarly the anticholinergic are avoided because they will relax the LES and thus GERD will increase.

    4) H2 antagonists are recommended because they decrease the acid secretion especially at night by blocking H2 receptors.

    5) Elevation of the head of the bed recommended in order to encourage the gravitational flows of the contents in stomach toward the pyloric end.

    6) Normal Stomach pH = 3-5

       Normal esophageal pH = 6-7

       pH at Lower esophagus  in gastroesophageal patient = 3-5

       pH of stomach in patient of GERD = 3-5 (same as that in normal)

  5. hi there

    Explanation:

    your answer

    when a object get physical force exerted on an object

    extra knowledge-basic formula is f/a(force per unit area )

    thanks you for brainliest follow me

  6. ( 1 ) The 26years old business executive is suffering from a disorder known as HIATAL HERNIA which is a structural defect in which a weakened diaphram allows a portion of the stomach to pass through the esophageal opening into the chest thus causing an increase in pressure.

    ( 2 ) The mechanisms that normally prevent gastric reflux into the esophagus when lying down or bending over is adequate lower esophageal pressure at the lower esophageal sphincter.

    ( 3 ) The parasympathetic stimulation (PS) of the autonomic nervous system innervates the lower esophageal sphincter,so cholinergic agonists will then increase lower esophageal sphincter (LES) contractions preventing gastric reflux. Anticholinergic agents will decrease lower esophageal sphincter pressure.

    ( 4 ) histamine (H2) antagonists are recommended because they reduce gastric acidity by selectively blocking H2 receptors (which are mediators for gastric secretion).

    ( 5 ) Elevation of the head of the bead is recommended because it encourages gravitational flow of the gastric contents towards the pyloric end of the stomach.

    ( 6 ) The normal PH of the esophagus is 6-7,and the normal PH of the stomach is 3-5.

    For this patient,the lower esophageal PH will be 3-5 while the stomach PH will not change.

  7. B.

    Gentle, long term pressure

    Explanation: The goods or services demand will be moving in a slow gradual way as consumers needs of the said services increases from an arithmetic to a geometric progression as population arises.

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