Recent content by Falinox
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2 problems: potential erg & tension/friction
Thank you! My mistake on question 7 was I forgot to multiply the 9.8(0.68) by 2 when I moved the 1/2 from the left side to the right side of the equation. I also came up with the answer 4.76 m/s. On #5 I got the answer a = .57 m/s squared can someone verify this for me.- Falinox
- Post #3
- Forum: Introductory Physics Homework Help
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2 problems: potential erg & tension/friction
Homework Statement http://i754.photobucket.com/albums/xx183/reddiesel08/Question1.jpg http://i754.photobucket.com/albums/xx183/reddiesel08/Question2.jpg 2. The attempt at a solution http://i754.photobucket.com/albums/xx183/reddiesel08/Attempt2.jpg...- Falinox
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- Potential
- Replies: 3
- Forum: Introductory Physics Homework Help
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How Do You Solve Kinematics Problems Involving Quadratic Equations?
Thanks everyone! I was able to figure it out.- Falinox
- Post #4
- Forum: Introductory Physics Homework Help
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How Do You Solve Kinematics Problems Involving Quadratic Equations?
Homework Statement http://i754.photobucket.com/albums/xx183/reddiesel08/Phyque.jpg 2. The attempt at a solution ∆x = ∆x v₀ + .5(a)(t²) = v(t) .5(9.8)t² = 35t 4.9t² - 35t = 0 Using the quadratic formula I get t = 7.95 sec ∆x = v(t) 35(7.95) = 278.25m Need help setting up and...- Falinox
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- Homework Kinematics
- Replies: 3
- Forum: Introductory Physics Homework Help
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Verifying Efficiency in a Thermodynamic System: Q1 & Q2
ok so... .35 = 1 - [(Tc + 273)K / (Tc + 180 + 273)K] .35 = 1 - [(Tc + 273K) / (Tc + 453K)] -.65 = - [(Tc + 273K) / (Tc + 453K)] (divide by -1 on both sides) .65 = [(Tc + 273K) / (Tc + 453K)] .65(Tc + 453K) = Tc + 273K .65Tc + 294.45K = Tc + 273K 21.45K = .35Tc Tc =...- Falinox
- Post #3
- Forum: Introductory Physics Homework Help
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Verifying Efficiency in a Thermodynamic System: Q1 & Q2
Homework Statement http://i754.photobucket.com/albums/xx183/reddiesel08/Q2.jpg Homework Equations e = 1 - (Tc/Th) (Temp in Kelvins) The Attempt at a Solution e = 1 - (300k/520k) = .42 or 42% No because the engine can only perform 42 J of work for each 100 J of energy. Homework...- Falinox
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- Efficiency System Thermodynamic
- Replies: 3
- Forum: Introductory Physics Homework Help
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Solving Atwood's Machine: Find Accel & Time for 60 cm
Homework Statement An Atwood’s Machine has a solid cylindrical pulley with a mass of 80 g and a radius of 10 cm. A 1.1 kg mass is hung on one side and a 1.05 kg mass on the other side. Find the acceleration of the two masses, and the time for them to travel a distance of 60 cm. 2. The attempt...- Falinox
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- Machine Time
- Replies: 1
- Forum: Introductory Physics Homework Help
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Rotational Motion of a wheel turning
Thanks for the reply! ω²=ω₀²+2αθ Given: ω₀=0 rad/s, α=1.2 rad/s² Is θ my 60 rev? (maybe convert 60 rev to rad?)- Falinox
- Post #4
- Forum: Introductory Physics Homework Help
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Rotational Motion of a wheel turning
Homework Statement A wheel starting from rest accelerates at 1.2 rad/s² counterclockwise. A) How long does it take to turn through 60 revolutions? B)What is the rotational speed at that time? 2. The attempt at a solution I converted 1.2rad/s² to 1.885 rev/s² by dividing by 2πrad (2...- Falinox
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- Motion Rotational Rotational motion Turning Wheel
- Replies: 4
- Forum: Introductory Physics Homework Help
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How Deep Does a Ferry Sink with Added Weight?
Homework Statement A) A ferry boat has dimensions of 25 m by 10 m by 5 m and a mass of 50,000 kg. How far below the surface of the water will the bottom of the boat be? (ρ=1015 kg/m3) B) If 12 cars of mass 1500 kg each drive onto the ferry, how much lower does the ferry sink into the water...- Falinox
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- Principle
- Replies: 1
- Forum: Introductory Physics Homework Help