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2 years ago

In the water cycle ,matter moves towards of gravity during

1 answer:

3 0

In the water cycle, matter moves against the pull of gravity during infiltration and evaporation transpiration and infiltration precipitation and infiltration evaporation and transpiration

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Between ball B and ball C, ball ____ has LESS potential energy because ___

Citrus2011 [14]

Answer: b

Explanation:

Ec= (1/2)m × v^2

By the formula, you can see that the bigger the mass, the bigger the Cinetic Energy.

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2 years ago

Malika is writing an essay about the Sun. Below is the first paragraph in the essay. The Sun is an average-size star that is loc

djverab [1.8K]

The 109 Earth's the sun can hold over billion Earth's

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2 years ago

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You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

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2 years ago

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Ted is a high school student. He is very good at math, but he is not a very good English student. He likes all of his teachers;

scoray [572]

I think the correct answer is

D) Ted associated being asked a question with embarrassment.

Glad I could help, and good luck!

AnonymousGiantsFan

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3 years ago

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If you were to come back to our solar system in 6 billion years what might you expect to find

Mariulka [41]

Dead starts bursting new ones being born, maybe more dwarf planets

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2 years ago