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3 years ago

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The current interest in continental drift was started with the notion of sea- floor spreading, which was propounded by Hess and

Dietz in the ________.A) 1700s B) 1850s C) 1990s D) 1890s E) 1960s

1 answer:

timofeeve [1]3 years ago

7 0

Answer:

idk

Explanation:

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Technician a says multiple discharge ignition system fires the spark plug during each of the engine's four cycle strokes. Techni

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Technician B only is correct

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The dual spark plug ignition system is a method if obtaining the ideal combustion and improved fuel consumption than a single spark plug ignition system.

Therefore, only technician B is correct.

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3 years ago

The terrestrial planets and the giant planets have different compositions because

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The terrestrial planets are closer to the sun.

Explanation:

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An aurora occurs when ____

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The short answer to how the aurora happens is that energetic electrically charged particles (mostly electrons) accelerate along the magnetic field lines into the upper atmosphere, where they collide with gas atoms, causing the atoms to give off light.

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Calculate the total displacement of a mouse walking along a ruler, if it begins at the x=5cm, and then does the following: It wa

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3 years ago

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

5 0

3 years ago

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