PS 113 Final Exam Prep

34 flashcards · Physics · by Nicholas Enoul

  1. An arrow is shot from a height of 1.4 m toward a cliff of height H. It is shot with a velocity of 25 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.4s later (a). What is the height of the cliff? (b). What is the max height reached by the arrow along its trajectory? #3.8

    (a) 18.4 (b) 25.3 (an attempt at an explanation) Formulas to use: ---- For (a) ---- Since we're going to be working on the Y axis for the final H, we almost have everything for the formula except Ay and Voy. Ay will be our gravity, while we need to apply some trig to the velocity. Overall velocity = 25 m/s \\ As we are working with the Y axis, we will want to use sine (remember SohCahToa!) Vy = 25*Sin(60) Ay = -9.8 T = 3.4 Yo = 1.4 \\ We now want to plug all of this into the first formula! y =…

  2. An arrow is shot from a height of 1.7 m toward a cliff of height H. It is shot with a velocity of 29 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.6 s later (a). What is the height of the cliff? (b). What is the max height reached by the arrow along its trajectory? #3.8

    (a) 28.6 (b) 33.9 (an attempt at an explanation) Formulas to use: ---- For (a) ---- Since we're going to be working on the Y axis for the final H, we almost have everything for the formula except Ay and Voy. Ay will be our gravity, while we need to apply some trig to the velocity. Overall velocity = 25 m/s \\ As we are working with the Y axis, we will want to use sine (remember SohCahToa!) Vy = 29*Sin(60) Ay = -9.8 T = 3.6 Yo = 1.7 \\ We now want to plug all of this into the first formula! y =…

  3. A very strong, but inept, shot putter puts the shot straight up with an initial velocity of 10.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.70 m tall (he could end up shorter)?

    Answer: 2.09s (an attempt at an explanation) Formulas to use: \\ With all the units we have, we can directly use the first formula: Yf = 1.7 Yi = 2.2 Ay = -9.8 Viy = 10 T = what we're trying to find. \\ Plugged in, it should look like this: 1.7 = 2.2 + 10(t) + (1/2)(-9.8)(t^2) 0 = 2.2-1.7 + 10t -4.9t^2 \\ As we can see here, we can actually apply the quadratic formula so long as we don't mix up our variables. To do this, we could either use it as is (remembering that A starts from the right as…

  4. A very strong, but inept, shot putter puts the shot straight up with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.30 m, and he is 1.70 m tall (he could end up shorter)?

    Answer: 2.3s (an attempt at an explanation) Formulas to use: \\ With all the units we have, we can directly use the first formula: Yf = 1.7 Yi = 2.3 Ay = -9.8 Viy = 11 T = what we're trying to find. \\ Plugged in, it should look like this: 1.7 = 2.3 + 11(t) + (1/2)(-9.8)(t^2) 0 = 2.3-1.7 + 11t -4.9t^2 \\ As we can see here, we can actually apply the quadratic formula so long as we don't mix up our variables. To do this, we could either use it as is (remembering that A starts from the right as w…

  5. A projectile is launched at ground level with an initial speed of 48.0 m/s at an angle of 25.0° above the horizontal. It strikes a target above the ground 2.90 seconds later. What are the x and y distances from where the projectile was launched to where it lands? Question #: 3.6

    x = 126.2 y = 17.6 (an attempt at an explanation) This question must be broken up into two parts between the X and Y axis. -- For X -- We know our initial velocity is 48 m/s, however, this is the total velocity from both the Y and X axis. Since we are given the reference angle, we can use Cosine to figure out the X component of this initial velocity. Thusly: Vox = 48cos(25) T = 2.9 Xo = 0 (this is our starting point) Ax = 0 (On the X axis, the projectile will not be accelerating any further - o…

  6. A projectile is launched at ground level with an initial speed of 51.5 m/s at an angle of 25.0° above the horizontal. It strikes a target above the ground 2.60 seconds later. What are the x and y distances from where the projectile was launched to where it lands? Question #: 3.6

    x = 121 y = 23.5 (an attempt at an explanation) This question must be broken up into two parts between the X and Y axis. -- For X -- We know our initial velocity is 51.5 m/s, however, this is the total velocity from both the Y and X axis. Since we are given the reference angle, we can use Cosine to figure out the X component of this initial velocity. Thusly: Vox = 51.5cos(25) T = 2.6 Xo = 0 (this is our starting point) Ax = 0 (On the X axis, the projectile will not be accelerating any further -…

  7. A 4.85 ✕ 105 kg rocket is accelerating straight up. Its engines produce 1.650 ✕ 107 N of thrust, and air resistance is 4.60 ✕ 106 N. What is the rocket's acceleration (in m/s2)? (Enter the magnitude.) Simplified: 485,000kg weight 16,500,000 thrust (five zeros) 4,600,000 air resistance (five zeros) Question #: 4.7

    Answer: 14.7 On the surface, it looks like we have everything we need for F = MA. However, this is meant to TRICK YOU! They said air resistance only, NOT including gravity! As we are going straight up, we'll need to account for the weight from gravity! With this, use W = MG to find the weight, then subtract the thrust with this w. Once you have this, continue as normal. W=MG --> 485,000 (9.8) --> W = 4,753,000 F - Friction - Weight = MA 16,500,000 - 4,600,000 - 4,753,000 = 485,000A 7,147,000/48…

  8. A 5.05 ✕ 105 kg rocket is accelerating straight up. Its engines produce 1.300 ✕ 107 N of thrust, and air resistance is 4.25 ✕ 106 N. What is the rocket's acceleration (in m/s2)? (Enter the magnitude.) Simplified: 505,000kg mass 13,000,000 thrust (six zeros) 4,250,000 air resistance (four zeros) Question #: 4.7

    Answer: 7.53 (an attempt at an explanation) On the surface, it looks like we have everything we need for F = MA. However, this is meant to TRICK YOU! They said air resistance only, NOT including gravity! As we are going straight up, we'll need to account for the weight from gravity! With this, use W = MG to find the weight, then subtract the thrust with this w. Once you have this, continue as normal. W=MG --> 505,000(9.8) --> W = 4,949,000 F = MA F - Friction - Weight = MA 13,000,000 - 4,250,00…

  9. Calculate the speed (in m/s) a spherical raindrop would achieve falling from 2.70 km in the absence of air drag and with air drag. Take the size across of the drop to be 8 mm, the density to be 1.00 ✕ 103 kg/m3, and the surface area to be 𝜋r2. (Assume the density of air is 1.21 kg/m3.) (a) in the absence of air drag (b) with airdrag NOTE: Coefficient of a sphere is 0.45. #5.10

    (a) 230 m/s (b) 14 m/s (an attempt at an explanation) -- For (a) -- This part is super simple as we don't actually need to use any of the formulas from this chapter. Instead, we will need to use the following from chapter 3/4: With this, the only thing we need to do now is to convert one unit then input into the formula. 2.7km = 2,700 m ay = -9.8 Vfy^2 = 0^2 + 2 (-9.8) (-2,700) <-- negative just so that we can still root afterwards. Vfy = sqroot(52920) Vfy = ~230 m/s -- For (b) -- For this one,…

  10. Calculate the speed (in m/s) a spherical raindrop would achieve falling from 1.40 km in the absence of air drag and with air drag. Take the size across of the drop to be 2 mm, the density to be 1.00 ✕ 103 kg/m3, and the surface area to be 𝜋r2. (Assume the density of air is 1.21 kg/m3.) (a) in the absence of air drag (b) with airdrag NOTE: Coefficient of a sphere is 0.45. #5.10

    (a) 166 m/s (b) 6.93 m/s (an attempt at an explanation) -- For (a) -- This part is super simple as we don't actually need to use any of the formulas from this chapter. Instead, we will need to use the following from chapter 3/4: With this, the only thing we need to do now is to convert one unit then input into the formula. 1.4km = 1,400 m ay = -9.8 Vfy^2 = 0^2 + 2 (-9.8) (-1,400) <-- negative just so that we can still root afterwards. Vfy = sqroot(27440) Vfy = ~166 m/s -- For (b) -- For this on…

  11. What is the ideal banking angle (in degrees) for a gentle turn of 1.05 km radius on a highway with a 120 km/h speed limit (about 75 mi/h), assuming everyone travels at the limit? DANGER: my calculator kept getting it wrong. Don't do, for example, v^2/100*8. Instead, do vfinal/800 --> this is a sure-fire way of getting the right answer. Question #: 6.7

    Answer: 6.16 (an attempt at an explanation) Other than a formula, this is mainly a unit conversion question. The unit we want is m or m/s. r = 1.05 km (convert KM to M) * 1000m r = 1050 meters v = 120 km/h (convert to m/s) * 1000m / 3600s (in an hour) v = ~33.33 m/s Once converted, we'll want to use the following formula: tan(theta) = (v^2)/rg plug this stuff in, and we're almost there! tan(theta) = (33.33^2)/1050(9.8) tan(theta) = 1110.8889/10290 tan(theta) = ~0.108 then we use inverse tangent…

  12. What is the ideal banking angle (in degrees) for a gentle turn of 1.15 km radius on a highway with a 115 km/h speed limit (about 71 mi/h), assuming everyone travels at the limit? DANGER: my calculator kept getting it wrong. Don't do, for example, v^2/100*8. Instead, do vfinal/800 --> this is a sure-fire way of getting the right answer. Question #: 6.7

    Answer: 5.17 (an attempt at an explanation) Other than a formula, this is mainly a unit conversion question. The unit we want is m or m/s. r = 1.15 km (convert KM to M) * 1000m r = 1150 meters v = 115 km/h (convert to m/s) * 1000m / 3600s (in an hour) v = ~31.94m/s Once converted, we'll want to use the following formula: tan(theta) = (v^2)/rg plug this stuff in, and we're almost there! tan(theta) = (31.94^2)/1150(9.8) tan(theta) = 1020.1636/11270 tan(theta) = ~0.09052 then we use inverse tangen…

  13. During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 67.0 kg and its horizontal component of velocity is 9.00 m/s when the 65.0 kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity in meters per second? #8.10 https://www.youtube.com/watch?v=cR92AwexbBg

    Answer: 4.57 m/s (an attempt at an explanation) This may seem daunting, but it is a super easy plug n' play question with one funky aspect. Here's the steps: We will have to use the following formula: \\ With this, we have our variables: M1 = 67 V1i = 9 M2 = 65 V2i = 0 <-- This is because we're assuming the performer is stationary! V2f and V1f = What we are trying to find. \\ With this first part done, we can now plug stuff in, which should look like: (67)(9) + (65)(0) = (67)V1f + (65)V2f \\ Th…

  14. During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 54.0 kg and its horizontal component of velocity is 7.50 m/s when the 65.0 kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity in meters per second? #8.10 https://www.youtube.com/watch?v=cR92AwexbBg

    Answer: 3.4 m/s (an attempt at an explanation) This may seem daunting, but it is a super easy plug n' play question with one funky aspect. Here are the steps: We will have to use the following formula: \\With this, we have our variables: M1 = 54 V1i = 7.5 M2 = 65 V2i = 0 <-- This is because we're assuming the performer is stationary! V2f and V1f = What we are trying to find. \\With this first part done, we can now plug stuff in, which should look like: (54)(7.5) + (65)(0) = (54)V1f + (65)V2f \\…

  15. Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 120,000 kg and a velocity of 0.300 m/s, and the second having a mass of 95,000 kg and a velocity of −0.120 m/s. (The minus indicates direction of motion.) What is their final velocity (in m/s)? #8.5 https://www.youtube.com/watch?v=UdN6nBXdpy0

    Answer: 0.1144 m/s (an attempt at an explanation) This may seem daunting, but it is a super easy plug n' play question with one funky aspect. Here's the steps: We will have to use the following formula: \\With this, we have our variables: M1 = 120,000 V1i = 0.3 M2 = 95,000 V2i = -0.12 V2f and V1f = What we are trying to find. \\With this first part done, we can now plug stuff in, which should look like: (120,000)(0.3) + (95,000)(-.12) = (120,000)V1f + (95,000)V2f \\ As the train cars are couple…

  16. Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 180,000 kg and a velocity of 0.300 m/s, and the second having a mass of 95,000 kg and a velocity of −0.120 m/s. (The minus indicates direction of motion.) What is their final velocity (in m/s)? #8.5 https://www.youtube.com/watch?v=UdN6nBXdpy0

    Answer: 0.155 m/s (an attempt at an explanation) This may seem daunting, but it is a super easy plug n' play question with one funky aspect. Here's the steps: We will have to use the following formula: \\With this, we have our variables: M1 = 180,000 V1i = 0.3 M2 = 95,000 V2i = -0.12 V2f and V1f = What we are trying to find. \\With this first part done, we can now plug stuff in, which should look like: (180,000)(0.3) + (95,000)(-.12) = (120,000)V1f + (95,000)V2f \\ As the train cars are coupled…

  17. Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 6000 kg satellite uses this method to separate from the 1600 kg remains of its launcher, and that 8000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities (in m/s) using the frame of reference in which they were at rest before separation? (Assume the final velocity of the satellite is in the positive direction. Indicate the direction with the sig…

    Velocity of Satellite = 0.749 Velocity of Launcher = -2.81 (an attempt at an explanation) This seems stupidly daunting, but let's break it down step by step. The first thing to do is to use the formula: which then transforms to: \\ As the initial velocity is zero, the whole left side is canceled. Leaving us with this: \\ The next step is to start labeling. This way, we can find out what we want. Mass and Velocity of Launcher = ML & VL Mass and Velocity of Satellite = VS & VS \\ With this, it sh…

  18. Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 5200 kg satellite uses this method to separate from the 1200 kg remains of its launcher, and that 3000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities (in m/s) using the frame of reference in which they were at rest before separation? (Assume the final velocity of the satellite is in the positive direction. Indicate the direction with the sig…

    Velocity of Satellite = 0.466 Velocity of Launcher = -2.015 (an attempt at an explanation) This seems stupidly daunting, but let's break it down step by step. The first thing to do is to use the formula: which then transforms to: \\ As the initial velocity is zero, the whole left side is canceled. Leaving us with this: \\ The next step is to start labeling. This way, we can find out what we want. Mass and Velocity of Launcher = ML & VL Mass and Velocity of Satellite = MS & VS \\ With this, it s…

  19. A 60.0 kg skier with an initial speed of 13.0 m/s coasts up a 2.50 m high rise as shown in the following figure. Question: Find her final speed at the top (in m/s), given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) #7.6 NOTE: I needed to use brute-force memorization for parts of this question. They will be on the answers card if you want to know them. Video Guide…

    Answer: 10.7 (an attempt at an explanation) Please note that I needed to use brute-force memorization for two particular things: \\ It is highly unlikely that, at the very least, the first one would be on the equations sheet, as we were meant to use the base Wnc = FDcos(theta), but by God I could not explain how or why this gets from that to -muse bullcrap. PLEASE DO EXTRA RESEARCH if you want to guarantee a 100%. Now, let's actually start on the first thing we need to do. We will need to use t…

  20. A 65.0 kg skier with an initial speed of 12.0 m/s coasts up a 2.50 m high rise as shown in the following figure. Question: Find her final speed at the top (in m/s), given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) #7.6 NOTE: I needed to use brute-force memorization for parts of this question. They will be on the answers card if you want to know them. Video Guide…

    Answer: 9.46 Margin of .2 (an attempt at an explanation) Please note that I needed to use brute-force memorization for two particular things: \\ It is highly unlikely that, at the very least, the first one would be on the equations sheet, as we were meant to use the base Wnc = FDcos(theta), but by God I could not explain how or why this gets from that to -muse bullcrap. PLEASE DO EXTRA RESEARCH if you want to guarantee a 100%. Now, let's actually start on the first thing we need to do. We will…

  21. (a) How high a hill can a car coast up (engine disengaged) if friction is negligible and its initial speed is 84.0 km/h? (b) If, in actuality, a 750 kg car with an initial speed of 84.0 km/h is observed to coast up a hill to a height 13.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope 2.5° above the horizontal? NOTE: I am not sure about this, but you may need to brute-force memorize the formula for…

    (a) 27.8 (b) 108616.67 or 1.09e+05 (c) 364 newtons (an attempt at an explanation) First thing to do: Convert the units we have to MKS. 84 km/h * 1000m/km * 1hr/3600 seconds 84 km/hr --> ~23.33 m/s --- Solving for (a) --- We will first need the following formulas or (a): \\ Please note that the final Kinetic force (Kf) and PEi are both 0. Why? Idk, I ain't paid to do this! However, with all of this info, we can make the following formula: \\ With this, we can now solve for Y, and fun fact, we do…

  22. (a) How high a hill can a car coast up (engine disengaged) if friction is negligible and its initial speed is 108.0 km/h? (b) If, in actuality, a 750 kg car with an initial speed of 108 km/h is observed to coast up a hill to a height 24.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope 2.5° above the horizontal? NOTE: I am not sure about this, but you may need to brute-force memorize the formula for…

    (a) 45.9 (b) 161100 or 1.61e+05 (c) 293 newtons (an attempt at an explanation) First thing to do: Convert the units we have to MKS. 108 km/h * 1000m/km * 1hr/3600 seconds 108 km/hr --> 30 m/s --- Solving for (a) --- We will first need the following formulas or (a): \\ Please note that the final Kinetic force (Kf) and PEi are both 0. Why? Idk, I ain't paid to do this! However, with all of this info, we can make the following formula: \\ With this, we can now solve for Y, and fun fact, we don't n…

  23. (a) How long (in s) will it take an 845 kg car with a useful power output of 44.0 hp (1 hp = 746 W) to reach a speed of 17.0 m/s, neglecting friction? (Assume the car starts from rest.) (b) How long (in s) will this acceleration take if the car also climbs a 2.60 m-high hill in the process? #7.9

    (a) 3.72 (b) 4.38 (an attempt at an explanation) --- Solving for (a) --- \\The first formula we would want to use is this one: \\ The good news is that we have P and we have the t to solve. Bad news: we don't know W yet. It is super easy though, use this formula: \\ This looks bad, but the great news is that Ki (initial kinetic energy) is completely gone as we are starting from rest. This leaves the following formula: \\ With this, we can solve for W M = 845 V = 17 (1/2) *845*17^2 --> W = 12210…

  24. (a) How long (in s) will it take an 875 kg car with a useful power output of 42.0 hp (1 hp = 746 W) to reach a speed of 13.0 m/s, neglecting friction? (Assume the car starts from rest.) (b) How long (in s) will this acceleration take if the car also climbs a 2.80 m-high hill in the process? #7.9

    (a) 2.36 (b) 3.13 (an attempt at an explanation) --- Solving for (a) --- \\The first formula we would want to use is this one: \\ The good news is that we have P and we have the t to solve. Bad news: we don't know W yet. It is super easy though, use this formula: \\ This looks bad, but the great news is that Ki (initial kinetic energy) is completely gone as we are starting from rest. This leaves the following formula: \\ With this, we can solve for W M = 875 V = 13 (1/2) *875*13^2 --> W = 73937…

  25. (a) What is the mechanical advantage of a nail puller—similar to the one shown in the figure below—where you exert a force 30 cm from the pivot and the nail is 1.2 cm on the other side? (b) What minimum force (in N) must you exert to apply a force of 1225 N to the nail? (Enter the magnitude.) #9.8 https://www.youtube.com/watch?v=A8nZgN3P3SY

    (a) 25 (b) 49 (an attempt at an explanation) The only formula(s) you will be needing to use for both these problems is the following: ---- For (a) ---- \\ With the formula above, we can input the numbers we have directly as: Formula: ME = Li/Lo (because Fo/Fi = Li/Lo) Li = 30cm <--- This is because we are exerting the force from the pivot (aka our input distance) Lo = 1.2cm ME = 30/1.2 <--- Don't need to worry about units bc they cancel out ME = 25 ---- For (b) ---- \\ We use the exact same for…

  26. (a) What is the mechanical advantage of a nail puller—similar to the one shown in the figure below—where you exert a force 50 cm from the pivot and the nail is 1.6 cm on the other side? (b) What minimum force (in N) must you exert to apply a force of 1500 N to the nail? (Enter the magnitude.) #9.8 https://www.youtube.com/watch?v=A8nZgN3P3SY

    (a) 31.3 (b) 48 (an attempt at an explanation) The only formula(s) you will be needing to use for both these problems is the following: ---- For (a) ---- \\ With the formula above, we can input the numbers we have directly as: Formula: ME = Li/Lo (because Fo/Fi = Li/Lo) Li = 50cm <--- This is because we are exerting the force from the pivot Lo = 1.6cm ME = 50/1.6 ME = 31.25 ---- For (b) ---- \\ We use the exact same formula, but now we are inputting 1225 as our Fo (applying the force to the nai…

  27. Suppose you needed to raise a 295 kg mower a distance of 6.0 cm above the ground to change a tire. If you had a 1.7 m long lever, where would you place the fulcrum (in m) if your force was limited to 390 N? (Enter the minimum distance measured from your end of the lever.) https://www.youtube.com/watch?v=pn4JLMx0Lss #9.9

    Answer: 1.5 (an attempt at an explanation) \\The first step is to figure out what formula to use. For this one, the following should be used: \\From here, there is one last thing we should do. Since the total length of the lever is 1.7m long, that must mean that the sum total of 1.7 meters \\ For reasons that will be apparent soon, we will want to do some basic algebra to substitute Lo with (this is because in the first formula, we want the denominator to be substituted so that we can cross mul…

  28. Suppose you needed to raise a 205 kg mower a distance of 6.0 cm above the ground to change a tire. If you had a 1.2 m long lever, where would you place the fulcrum (in m) if your force was limited to 320 N? (Enter the minimum distance measured from your end of the lever.) https://www.youtube.com/watch?v=pn4JLMx0Lss #9.9

    Answer: 1.04 (an attempt at an explanation) \\ The 6.0 cm is a scam, you don't need to ever worry about that! \\The first step is to figure out what formula to use. For this one, the following should be used: \\From here, there is one last thing we should do. Since the total length of the lever is 1.7m long, that must mean that the sum total of 1.7 meters \\ For reasons that will be apparent soon, we will want to do some basic algebra to substitute Lo with (this is because in the first formula,…

  29. What is the final velocity (in m/s) of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest? https://www.youtube.com/watch?v=r1-gKqMM8AA&t= #10.5

    Answer: 7 (an attempt at an explanation) BIG HEADS UP: There are two (or one and a half) ways to solve this problem. Either we can solve it normally, or cheese through it completely. If you want the cheese, simply use the following formula: Final Velocity = sqroot(9.8*height) If you only memorize this "formula", you might not be able to answer the first question on the quiz (what formula to use) since this is the end result of a lot of algebra and physics theory. If you want a 100% chance of so…

  30. What is the final velocity (in m/s) of a hoop that rolls without slipping down a 3.50-m-high hill, starting from rest? https://www.youtube.com/watch?v=r1-gKqMM8AA&t= #10.5

    Answer: 5.86 (an attempt at an explanation) BIG HEADS UP: There are two (or one and a half) ways to solve this problem. Either we can solve it normally, or cheese through it completely. If you want the cheese, simply use the following formula: Final Velocity = sqroot(9.8*height) If you only memorize this "formula", you might not be able to answer the first question on the quiz (what formula to use) since this is the end result of a lot of algebra and physics theory. If you want a 100% chance of…

  31. A ball with an initial velocity of 7.33 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calculate the vertical height (in m) it reaches. (b) Repeat the calculation (in m) for the same ball if it slides up the hill without rolling. https://www.youtube.com/watch?v=0_FbQtzuLSU #10.6

    (a) 4.57 (b) 2.74 (an attempt at an explanation) ---- For (a) ---- The very first thing we want to do is use the following formula. I'm not entirely sure how we got there, but it is simple to memorize and relatively similar to chapter 7 (plz watch the video to fully understand): \\ After this, we want to break the KE and PE down into their components (same as from Quiz 2 and Chapter 7). The only difference being KEri having it's own special thing (with "I" and "Omega"). \\ Quick note, after thi…

  32. A ball with an initial velocity of 8.11 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calculate the vertical height (in m) it reaches. (b) Repeat the calculation (in m) for the same ball if it slides up the hill without rolling. https://www.youtube.com/watch?v=0_FbQtzuLSU #10.6

    (a) 5.59 (b) 3.36 (an attempt at an explanation) ---- For (a) ---- The very first thing we want to do is use the following formula. I'm not entirely sure how we got there, but it is simple to memorize and relatively similar to chapter 7 (plz watch the video to fully understand): \\ After this, we want to break the KE and PE down into their components (same as from Quiz 2 and Chapter 7). The only difference being KEri having it's own special thing (with "I" and "Omega"). \\ Quick note, after thi…

  33. (a) What force (in N) must be exerted by the wind to support a 2.40 kg chicken in the position shown in the figure? (Assume r⊥ = 21.0 cm and r'⊥ = 8.20 cm.) (b) What is the ratio of this force to the chicken's weight? (c) Does this support the contention that the chicken has a relatively stable construction? (Yes or No?) https://www.youtube.com/watch?v=gjlPAWNcFwI&t= #9.6

    (a) 9.18N (b) 0.39 (c) Yes (an attempt at an explanation) NOTE: I don't know what the hecc I am doing, nor do I entirely know the theory behind this. If you want a good explanation please review the video and visit office hours. ---- For (a) ---- This is all gonna be guesses (not even educated guesses, just shots in the dark), but here goes: \\ We wan't to start with this formula: \\ We now want to break down our ∑τ into it's components. There are two main forces at play: the chickens weight an…

  34. (a) What force (in N) must be exerted by the wind to support a 2.00 kg chicken in the position shown in the figure? (Assume r⊥ = 17.0 cm and r'⊥ = 9.20 cm.) (b) What is the ratio of this force to the chicken's weight? (c) Does this support the contention that the chicken has a relatively stable construction? (Yes or No?) https://www.youtube.com/watch?v=gjlPAWNcFwI&t= #9.6

    (a) 10.61 (b) 0.541 (c) Yes (an attempt at an explanation) NOTE: I don't know what the hecc I am doing, nor do I entirely know the theory behind this. If you want a good explanation please review the video and visit office hours. ---- For (a) ---- This is all gonna be guesses (not even educated guesses, just shots in the dark), but here goes: \\ We wan't to start with this formula: \\ We now want to break down our ∑τ into it's components. There are two main forces at play: the chickens weight a…