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Final_HB

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Everything posted by Final_HB

  1. ya looking back i never swapped signs after bring u2 across, so I missed it there Thanks for your help
  2. i know... I do have it somewhere in my head... just I need someone to talk me through it sometimes thanks the original should be sin2a and sin2b not just sin a and sin b as i have :/ and the wrong signs are still an issue, but it might just be my algebra
  3. Perfect... Just wanted to be sure I was on the right track with it before doing more work. Move some stuff around in both to get it equal to u2 so both equations become: u2=gsin2C-gsin2a and u2= gsin2C-gsin2b D1 D2 Let them equal each other. g cancels. remove fraction and you get: D1Sin2C-D1sin2b=D2sin2C-D2Sin2a And from there it pops out after some factorising EDIT:i actually ended with minuses in instead of pluses :/
  4. Doing that we get: u2Sin2a +D1 = u2Sin2b - D2 =u2SinC g g g ->u2(Sin2a-Sin2b) D2+D1=u2SinC g g
  5. Sorry... R= u2Sin2(theta) is the range formula the general one anyways. g Would you be able to say that the angle C would be between a and b?
  6. Okay then Inital launch speed is u. Time of flight can be found using one of the main physics formulas, which comes out as: T=2uSina g g=acceleration due to gravity And our range for the first lunch comes out as: D1= 2u2Cos(a)sin(a) = 2u2Sin(2a) g g -The work is the exact same for D2, except at angle b so D2 = 2u2Sin2b g We could do this again, with angle C. Would the range be D2-D1 ?
  7. When launched at an angle a to the horizontal, a projectile falls a distance D1 short of a target. If the projectile is fired at an angle b to the horiztonal, it reaches a distance D2 beyond the target. Assuming that C is the correct angle for a launched projectile to hit the target, show that: Sin2C = D2Sina + D1Sinb D1+D2 Well... If its a 'Sin2C' identity, it probably comes from Cosine times Sine. It probably goes like: -Find time of flight -Find a formula expressing the launch angle. -rinse and repeat for the second launch. -Equate the two, and mess with it til the desired result. EDIT: question states that both launch speeds are the same.
  8. Cause the sum is 0, they can be just taken out of the formula and so F1+F2=0 is the only forces needed for equilibruim
  9. Fair enough... The sum of the three forces is 0.
  10. Sorry, I may have been unclear. I think I was referring to the same thing you are... But you used the correct terminology
  11. Thank you Just making sure before I progress through it
  12. I think so. something like: F1+F2+F3+F4+F5=0 And when its reversed: F1+F2-F3-F4-F5=0 let the two equal each other and cancel?
  13. A particle is under the influence of 5 forces. Three of these forces are reversed, and the particle remains in equilibrium. Prove that the particle will remain in equilibrium even if these three forces were removed altogether. My thinking for this is: With three of the forces gone, the 2 forces (F1 and F2 ) left must balance the system if its to remain at equilibrium. If change of forces occurs, the system will react by moving to a new state of equilibrium. No reaction means that the 3 forces in the question cancel out each other, and no shift occurs. If the three forces balance to 0, there is no need to have them as acting on the particle. Right? If im right, Is there any maths way to say this, or a better way to explain what im trying to say. Thank you in advance.
  14. A force of magnitude F acting up and along a smooth inclined plane, can support a mass M in equilibrium. If a force of the same magnitude acts horizontally, it can support a mass m on the same inclined plane in equilibrium. Find a relationship between F, M and m which is independent of the angle of inclination of the slope. Im not so sure on this one... I guess you: Take each force and particle separately. Resolve the forces into horizontal/vertical components. Let the sum of all these forces equal 0. Do the same for the second particle and equate the two equations to each other through F. Good so far?
  15. Im glad you enjoy them I might have to spam a few more here
  16. Hello again Sounds, and looks good to me This place is great so helpful, and quick to respond as well!! thank you again.
  17. Original question: A load of weight W is supported in equilibrium by two strings attached to it. One is inclined at an angle 30 to the vertical, and the other is inclined at an angle 60 to the vertical. If either rope breaks if the tension exceeds a value of To, Find the greatest value of W that can be supported. What I have so far: Well, its at equilibruim. So all the forces equal 0. Since we are looking for the greatest value, we set the tension in both strings as To as specified. Its at angles, so we need to get the horizontal and vertical components of each string. Which look like: To Cos30 + To Sin30 and To Cos60 +To Sin60 Am I good so far? Main problem is working out the tension... So hints on where to go from there would be greatly appreciated
  18. Hey just saw the post. Wow, okay that works to Im just used to splitting the graphs into easy geometric shapes... squares, retangles, triangles ya know... Because thats how we were shown In all honesty, I didnt realize that there an area formula for traps Thank you, imatfaal and studiot
  19. Thank you!! Agreed :/ thanks all the same
  20. The original post has the question word for word Ta +Td +Tc =240 From what I said in Post #5, 2Ta + Tc =240 So... simultaneous equations and solve for Tc ?
  21. Sorry. The area of the triangles is the distance travelled while accelerating/decelerating. Uniform... Does that mean they are equal, or constant? If they are equal, the distance travelled must be the same... so Ta = Td
  22. I assumed I did Yup... drawn. Wouldnt get me an A in art, but its here. The graph divides up into three shapes, Two triangles, and a retangle. The retangle is the area we are looking for, and the triangles are the acceleration/ deceleration parts. EDIT: CallingTa the time accelerating and Td the time decelerating, we can equate the distance to the following: 60Ta/2 + 60Td/2 + 60tc =9000 which reduces to, Ta +Td +2Tc =300
  23. Hey. This is a little tricky The question goes: A train travels from rest at station A to rest again at station B in a time of 4 minutes and a distance of 9km. The greatest speed attained is 60m/s, and both the acceleration and decceleration are uniform. Find how far the train goes at full speed? I know what I need to do to get the answer, but I dont have enough info for it. -Find the time at a constant speed (Tc) -Find distance using the formula: S=UTc To find Tc, I need either: -Distance travelled during deceleration/acceleration. -Time to accelerate from rest to 60m/s. -Time to decelerate from 60m/s to rest. Get any of those four, some maths wizardry and im golden But i cant do anything with the info given. Am I missing a beat?
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