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emmasim [6.3K]
3 years ago
5

A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between t

he rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy?
(A) 3/7
(B) None of these
(C) 7/2
(D) 2/5
(E) 2/7
(F) 3/5
(G) 5/3
Physics
1 answer:
dsp733 years ago
4 0

Answer:

option E

Explanation:

given,

Rotational Kinetic Energy, KE_r = \dfrac{1}{2}I\omega^2

Moment of inertia of the solid,

I = \dfrac{2}{5}MR^2

\omega = \dfrac{V}{R}

now,

KE_r = \dfrac{1}{2}I\omega^2

KE_r = \dfrac{1}{2}\times \dfrac{2}{5}MR^2 (\dfrac{V}{R})^2

KE_r =\dfrac{1}{5}MV^2......(1)

transnational kinetic energy

KE_t =\dfrac{1}{2}MV^2

Total kinetic energy

KE = \dfrac{1}{2}MV^2 + \dfrac{1}{5} MV^2

KE = \dfrac{7}{10}MV^2

ratio of rotational kinetic energy to the total kinetic energy

\dfrac{KE_r}{KE_t}=\dfrac{\dfrac{1}{5}MV^2}{\dfrac{7}{10}MV^2}

\dfrac{KE_r}{KE_t}=\dfrac{2}{7}

hence, the correct answer is option E

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Answer:

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Substituting the values in the above equation

                                        13.33 = v²/30

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As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile
Ber [7]

Answer:

a)\Delta T= 100^{\circ}C

b)\bigtriangledown T=1^{\circ}C.mile^{-1}

c)\bigtriangledown T_4=1^{\circ}C.mile^{-1}

d)\bigtriangledown T_4=1^{\circ}C.mile^{-1}

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

T=(d+30) ^{\circ}C...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

\Delta T= T_f-T_i..............................(2)

where:

T_f= final temperature

T_i= initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

\Delta T= (100+30)-(0+30)

\Delta T= 100^{\circ}C

(b)

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\bigtriangledown T=\frac{\Delta T}{\Delta d}......................(3)

where:

\Delta d= change in distance

\bigtriangledown T=change in temperature with respect to distance

putting the respective values in eq. (3)

\bigtriangledown T=\frac{100}{100}

\bigtriangledown T=1^{\circ}C.mile^{-1}

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comparing the given function of the temperature with the general equation of  a straight line:

y=m.x+c

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

\bigtriangledown T_4=1^{\circ}C.mile^{-1}

(d)

comparing the given function of the temperature with the general equation of  a straight line:

y=m.x+c

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

\bigtriangledown T_4=1^{\circ}C.mile^{-1}

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