Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Mechanical Engineering homework question answer, step 1, image 1 Mechanical Engineering homework question answer, step 1, image 2 Step 2 Mechanical Engineering homework question answer, step 2, image 1 Step 3 Mechanical Engineering homework question answer, step 3, image 1

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Textbook Solution:

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Answer:

To determine

Rank the energy quantities from largest to smallest.

Answer

The ranking is a>b=c−−−−−−−−$\underset{\_}{a>b=c}$ the energy quantities from largest to smallest.

Explanation

a$a$ The absolute value of the average potential energy of the sun-earth system.

b$b$ The average kinetic energy of the earth in its orbital motion relative to the sun.

c$c$ The absolute value of the total energy of the sun-earth system.

Write the expression for the potential energy between the sun-earth system.

    U=−GMSMER$U=-\frac{G{M}_S{M}_E}{R}$        I$I$

Here, U$U$ is the average potential energy of the sun-earth system, G$G$ is the gravitational constant, MS$M_S$ is the mass of sun, ME$M_E$ is the mass of earth, R$R$ is the separation between the sun and earth.

Use equation I$I$ to write the absolute value of U$U$.

    |U|=GMSMER$\left|U\right|=\frac{G{M}_S{M}_E}{R}$        II$II$

Write the expression for the kinetic energy.

    KE=12MEv2E$K_E=\frac{1}{2}{M}_E{v}_E^2$        III$III$

Here, KE$K_E$ is the kinetic energy of the earth, vE$v_E$ is the velocity of the earth.

Write the expression for the gravitational force.

    FG=GMEMSR2$F_G=\frac{G{M}_E{M}_S}{R^2}$        IV$IV$

Here, FG$F_G$ is the gravitational force.

Write the expression for the centripetal force of the earth.

    FC=MEv2ER$F_C=\frac{M_E{v}_E^2}{R}$        V$V$

Here, FC$F_C$ is the centripetal force.

Equate equation IV$IV$ and V$V$ to solve for v2E$v_E^2$.

    v2E=GMSR$v_E^2=\frac{G{M}_S}{R}$        VI$VI$

Use equation VI$VI$ in III$III$ to solve for KE$K_E$.

    KE=GMSME2R$K_E=\frac{G{M}_S{M}_E}{2R}$        VII$VII$

Write the expression for the total energy of the sun-earth system.

    E=KE+U$E=K_E+U$        VIII$VIII$

Here, E$E$ is the total energy of the earth-sun system.

Use equation I$I$ and VII$VII$ in VIII$VIII$ to solve for E$E$.

    E=GMSME2R−GMSMER=−GMSME2R$\begin{array}{c}E=\frac{G{M}_S{M}_E}{2R}-\frac{G{M}_S{M}_E}{R}\\ =-\frac{G{M}_S{M}_E}{2R}\end{array}$        IX$IX$

Use equation IX$IX$ and write the absolute value of E$E$.

    |E|=GMSME2R$\left|E\right|=\frac{G{M}_S{M}_E}{2R}$        X$X$

Conclusion:

The ranking of the energy quantities from largest to smallest is |U|>KE=|E|$\left|U\right|>K_E=\left|E\right|$.

Therefore, the ranking is a>b=c−−−−−−−−$\underset{\_}{a>b=c}$ the energy quantities from largest to smallest.