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About rasen58

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  1. Okay thank you for the corrections on all those points!
  2. Okay thanks, I guess the textbook gave the wrong answer.
  3. Given a 2.0 kg mass at rest on a horizontal surface at point zero. For 30.0 m, a constant horizontal force of 6 N is applied to the mass. For the first 15 m, the surface is frictionless. For the second 15 m, there is friction between the surface and the mass. The 6 N force continues but the mass slows to rest at the end of the 30 m. The coefficient of friction between the surface and the mass is _____. To solve this, I found the velocity at 15 m by first using F=ma to find that the acceleration for the first 15 m is 3 m/s^2. Then I used a kinematic equation to find that the velocity at 15 m is sqrt(90). So then, for the second 15 m, I drew a force diagram and saw that for the mass to decelerate to 0 in the exact same distance as it took to accelerate, then the net force must be the same magnitude but in the reverse direction to slow it down. So I thought that since it was previously just 6 N to the right, I thought the force of friction would have to be 12 N to the left so that the net force is 6 N to the left. Force of friction = coeff_fric * m * g So that means that 12 = coeff_fric * m * g Solving for coeff_fric, I got 0.6. But that is apparently wrong since it's supposed to be 0.3. But the only way to get 0.3 is if Force of friction = 6 N to the left. But I don't see why it should be 6 N instead of 12 N to the left.
  4. Find the speed of a 3 kg mass when it reaches the 10 m mark. The net force is in the same direction as the motion. Moving on straight line (rectilinear) I think I'd have to use net work = change in kinetic energy theorem here. Since the net work would be the area under the force position graph, I counted the boxes under the graph. So I tried just counting all the boxes and got (around?) 39 boxes. I then multipled that by 20 since each box is 20 N and got 780 N Since W = F *d = 780 N * 10 m = 7800 N*m So 7800 = .5 m v^2 Solving for v, I got 22.8 m/s. But the answer is actually 24. I think that's due to me not counting the correct number of boxes? or is there a different way or a better way to approximate how many boxes are under the graph? Thanks
  5. There are two equal masses of silver insulated from each other. Sample A is maintained at 20 degrees C. Sample B has twice as much thermal energy as Sample A. Silver has a specific heat of 233 J/g*C. What is the temperature of sample B? I don't have any idea how to do this and don't have that good of a grasp of the term 'thermal energy'. I think thermal energy is equivalent to total kinetic energy of all the particles, so maybe you could use the formulas for average kinetic energy, but I'm not sure how. q = cm dT KEavg = (1/2)mv2 = (3/2)KT
  6. Oh wow that makes sense. Thanks! So it would still be at 0 degrees C.
  7. So if the net heat is 0 q = heat gained by raising 3kg ice from -10oC to 0oC + heat gained by converting 3kg ice to water at 0oC + heat gained raising 3kg of water fom 0oC to +xoC degrees + Heat lost by lowering 10kg water from +10oC to +xoC = 0 q = (1/2)(3)(0 - (-10)) + (3)(80) + (1)(3)(x-0) + (1)(10)(x-10) = 0 But after I solved for x now, I get 11.9 degrees C, which doesn't make sense either since it's greater than 10.
  8. 3 kg of ice whose temperature is -10 C is dropped into 10 kg of water whose temperature is 10 C. How much of the ice melts when the mixture is brought to equilibrium? I wanted to find the equilibrium temperature first so I used q = cm T And c (water) = 1 kcal/kg C, c (ice) = 0.5 kcal/kg C So (0.5)(3)(x - (-10)) = (1)(10)(10 - x) That turns into 1.5x + 15 = 100 - 10x x = 7.4 C So I then determined how much energy outputted when lowering the temperature of the water is q = cm T = (1)(10)(10 - 7.4) = 26 kcal Then, to bring the ice to 0 C requires q = cm T = (0.5)(3)(0 - (-10)) = 15 kcal Then, to cause the phase change, I found the latent heat. Heat of fusion = 80 kcal/kg q = mL = (3 kg)(80 kcal/kg) = 240 kcal But 240 is so much higher than 26, so how is this possible. What am I doing wrong?
  9. Oh wow, I used the diameter instead of the radius. Thanks.
  10. A flask is closed tightly with a rubber stopper 3.00 cm in diameter. All the air in the flask is then removed. How much force is necessary to remove the stopper from the evacuated flask? (1 atm = 1.013 x 10^5 N/m^2) I have no idea how to do this problem. I think I'm supposed to use Bernoulli's equation, but I'm not sure how exactly to use it. I also tried using P = F/A So F = P*A = 1 x 10^5 x pi x (0.03)^2 = 283 But the answer is 71.6 N.
  11. Oh thanks, forgot about that, it works now
  12. When tripling the volume of a given weight of gas at 27 C, keeping the pressure constant, the temperature must be raised to? I used V1/T1 = V2/T2 and set V1 to 1 L 1/27 = 3/T2 T2 = 81 C. But the answer says it's 627 C. Is it wrong?
  13. Yep, just learned that it is perpendicular so it makes sense now.
  14. A child runs in a circular path with constant speed. Which one of the following quantities is constant and non-zero? A. Linear velocity B. Angular velocity C. Centripetal acceleration D. Angular acceleration E. Total acceleration I thought none of them would be constant since the direction of the vectors would always be changing when you move in a circle. But since I had to pick an answer, I thought it might be between linear velocity and angular velocity and picked linear velocity since there is a constant speed. But the answer is angular velocity. Why?
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