Recent content by euphoriax
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Elastic Potential Energy and velocity of spring
But before the mass is released it's still being compressed 0.3m, and then after it moves up 0.2m. I tried that anyway, and it wasn't the correct answer...- euphoriax
- Post #3
- Forum: Introductory Physics Homework Help
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Momentum elastic colliion in 2D
Homework Statement A hockey player of mass 8.0x10 kg was skating at a velocity of 7.5 m/s [E] when he accidentally hit a linesman who was just standing on the ice. The mass of the linesma is 9.0x10 kg and his velocity after the collision was 3.0 m/s [N30°E]. a) find the velocity of the hockey...- euphoriax
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- 2d Elastic Momentum
- Replies: 1
- Forum: Introductory Physics Homework Help
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Elastic Potential Energy and velocity of spring
Homework Statement A spring, having a force constant of 6.0x102 N/m, is held in a vertical position and compressed 0.30m. A 5.0 kg mass is then placed on top of the spring. THe mass is then releases. Neglecting air resistance and the mass of the spring, calculate: a) the velocity of the...- euphoriax
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- Elastic Elastic potential energy Energy Potential Potential energy Spring Velocity
- Replies: 3
- Forum: Introductory Physics Homework Help
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Projectile Time variable confusion
oh, I see. Thanks!- euphoriax
- Post #9
- Forum: Introductory Physics Homework Help
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Projectile Time variable confusion
Oh, I see. So if the question mentions the landing elevation, like the first one did with the plain, then i must use the trajectory equation and the quadratic formula, but if it doesn't state that and only to the apex, then i must find the time to the apex and then multiply by two to find the...- euphoriax
- Post #7
- Forum: Introductory Physics Homework Help
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Projectile Time variable confusion
yeah since it was only half the trajectory, i had to multiply the time i found for the first half by 2 to find the time for the whole thing so i can use it in the equation to find the horizontal range. but i wanted to know how come for one question you had to multiply the time by 2, and for the...- euphoriax
- Post #5
- Forum: Introductory Physics Homework Help
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Projectile Time variable confusion
But then sometimes you don't multiply it and just use the time you found from the vertical... So here's an example: A mortar shell is fired at 80m/s at an angle of 65 degrees to the horizontal and falls on a plain 50m below the level of the hill from which it was fired. What is the horizontal...- euphoriax
- Post #3
- Forum: Introductory Physics Homework Help
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Projectile Time variable confusion
Homework Statement When I'm solving problems, I'm told that the time between the vertical and horizontal components are the same. So that means that I would have to use the same time when calculating for both the vertical and horizontal components right? But then there are times when i have...- euphoriax
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- Confusion Projectile Time Variable
- Replies: 8
- Forum: Introductory Physics Homework Help
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What is the maximum range of a projectile gun with a speed of 315m/s?
Homework Statement A gun fires a shell with a speed of 315m/s. Neglecting the effects of air resistance, calculate the maximum range of this gunHomework Equations t=(vf-vi)/a v=d/t The Attempt at a Solution it doesn't give me an angle so I am kind of confused on how to solve this problem. I...- euphoriax
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- Gun Projectile
- Replies: 1
- Forum: Introductory Physics Homework Help
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What is the Force on Charge B Due to Charges A and C?
Homework Statement Three charges A, B and C with charges respectively +3.2 x 10-6 C and -7.8 x 10-6 C are placed as seen in the diagram. What is the force on B because of A and C? Homework Equations FE=kq1q2/r2 The Attempt at a Solution dAB = 2.5m dCB= 1.5m k= 9.0x109...- euphoriax
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- Charge Electric Electric field Field
- Replies: 1
- Forum: Introductory Physics Homework Help
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Calculating Energy of Photoelectrons Ejected from a Magnesium Surface
Oh, that's true. thank you!- euphoriax
- Post #3
- Forum: Introductory Physics Homework Help
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Calculating Energy of Photoelectrons Ejected from a Magnesium Surface
Homework Statement A magnesium surface has a work function of 3.68 eV. Electromagnetic waves with a 215 nm wavelength strikes the surface and eject photoelectrons. Calculate the energy of the photoelectrons in Joules and in electron volts. W= 3.68 eV = 5.96x10-19 J wavelength= 215 nm =...- euphoriax
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- Physics Quantum Quantum physics
- Replies: 2
- Forum: Introductory Physics Homework Help