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2 years ago

Use the table to answer the question.

Physics

1 answer:

QveST [7]2 years ago

6 0

Answer:

Mechanical Waves

Explanation:

I took the practice

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A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the

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A 0.10 kg ball of dough is thrown straight up into the air with an initial speed of 15 m/s.

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<span>Mass of the ball is m = 0.10kg Initial speed of the Ball v = 15m/s a. When the ball is at maximum height the velocity is 0 Momentum of ball = mass x velocity Momentum = 0.10kg x 0 = 0 b. Getting the maximum height, Using the conservation of energy equation KEinitial = mgh 1/2mVin^2 = mgh => h = v^2/2g h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m Applying the conservation of energy equation at halfway V^2 = 2gh V = square root of (2x9.8x5.96) => V = square root of (116.816) So the velocity at the half way V = 10.81 m/s Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>

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3 years ago

The force that keeps objects moving in a circle is called centrifugal force.<br> True<br> False

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3 years ago

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What is the definition of differentiation geology?

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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

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3 years ago

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