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nekit [7.7K]
3 years ago
10

A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 x 10^-3 kg∙m2 is suspended from the ceiling. A rope passes over

it with a 2.0-kg block attached to one end and a 4.0-kg block attached to the other. The rope does not slip on the pulley. At any instant after the blocks start moving the object with the greatest kinetic energy is:
Select one:

a. the heavier block

b. none (all three objects have the same kinetic energy)

c. none (all three objects have the same kinetic energy)

d. either block (the two blocks have the same kinetic energy)

e. the pulley
Physics
1 answer:
solmaris [256]3 years ago
3 0

Answer:

maximum kinetic energy is for Pulley

e) The pulley

Explanation:

Let 4 kg block is moving downwards with speed "v" so we can say that 2 kg block will move upwards with same speed "v"

Now we know that pulling is in pure rotational motion

so we will have

\omega = \frac{v}{R}

\omega = \frac{v}{0.03}

now kinetic energy of each is given as

For 4 kg block

K_1 = \frac{1}{2}(4)(v^2) = 2v^2

for 2 kg block

K_2 = \frac{1}{2}(2)(v^2) = v^2

For pulley

K_3 = \frac{1}{2}I\omega^2

K_3 = \frac{1}{2}(4.5\times 10^{-3})\frac{(v^2)}{0.03^2}

K_3 = 2.5 v^2

So maximum kinetic energy is for Pulley

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Answer:

X=92.49 m

Explanation:

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By putting the values

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James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than
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Answer:

0.12 K

Explanation:

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m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h =  c x ΔT

9.8 x 51 = 4190 x ΔT

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Answer:

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Explanation:

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At any  distance x from point A mass density

\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x

\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x

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dm =λ dx

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dI=\lambda x^2dx

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I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx

By integrating above we can find that

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x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}

x_c= \dfrac{5}{9}L

So mass moment of inertia I=\dfrac {7}{12}\lambda_ 0 L^3 and location of center of mass  x_c= \dfrac{5}{9}L

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