PS 113 - Midterm
26 flashcards · Physics · by Nicholas Enoul
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…
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 -…
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 3.6s later. (a) What is the height of the cliff (in m)? (b) What is the maximum height (in m) reached by the arrow along its trajectory? Question #: 3.8
(a) 28.6 (b) 33.9 (an attempt at an explanation) -- for (a) -- To start, the initial velocity given to us is the total velocity from both the X and Y axes. Since we are only working with our Y, and with the angle given to us (60 degrees), we can find the Y component of the arrow's velocity using Sine function. With this, we can make our list as such: Voy = 29sin(60) T = 3.6 Yo = 1.7 Ay = -9.8 (gravity) Formula to use: x = xo + VoxT + 1/2(ax)(T^2) -- for (b) -- The key thing to take away is that…
An arrow is shot from a height of 1.3 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 3.4s later. (a) What is the height of the cliff (in m)? (b) What is the maximum height (in m) reached by the arrow along its trajectory? Question #: 3.8
(a) 30 (b) 33.5 (an attempt at an explanation) -- for (a) -- To start, the initial velocity given to us is the total velocity from both the X and Y axes. Since we are only working with our Y, and with the angle given to us (60 degrees), we can find the Y component of the arrow's velocity using Sine function. With this, we can make our list as such: Voy = 29sin(60) T = 3.4 Yo = 1.3 Ay = -9.8 (gravity) Formula to use: x = xo + VoxT + 1/2(ax)(T^2) -- for (b) -- The key thing to take away is that w…
(a) An airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 17° south of east. Its direction of motion relative to the earth is 45.0° south of west, while its direction of travel relative to the air is 10° south of west. What is the airplane's speed relative to the air mass in meters per second?} (b) What is the airplane's speed relative to the Earth in m/s? !!MAKE SURE YOUR CALCULATOR IS IN DEGREE MODE!! https://www.youtube.com/watch?v=W_QDe2J7bD4 Question #: 3.11
(a) ~69.3 (margin of 68-70) (b) ~35.6 (margin of 34-36) (an attempt at an explanation) -- For (a) -- The very first thing that should be done is to create a free-body diagram with the units given. Note that if we aren't given what the jet stream is relative to, it will be relative to the earth (Velocity of Air to Earth --> VAE) It should look something like this: VPA --> Velocity of Plane relative to Air VPE --> V of Plane relative to Earth VAE --> V of Air relative to Earth mag --> magnitude o…
(a) An airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 28° south of east. Its direction of motion relative to the earth is 45.0° south of west, while its direction of travel relative to the air is 10° south of west. What is the airplane's speed relative to the air mass in meters per second?} (b) What is the airplane's speed relative to the Earth in m/s? !!MAKE SURE YOUR CALCULATOR IS IN DEGREE MODE!! https://www.youtube.com/watch?v=W_QDe2J7bD4 Question #: 3.11
(a) ~74.62 (my original answer) or 75 rounded (b) ~47.98 (an attempt at an explanation) -- For (a) -- The very first thing that should be done is to create a free-body diagram with the units given. Note that if we aren't given what the jet stream is relative to, it will be relative to the earth (Velocity of Air to Earth --> VAE) It should look something like this: VPA --> Velocity of Plane relative to Air VPE --> V of Plane relative to Earth VAE --> V of Air relative to Earth The color coding s…
A powerful motorcycle can produce an acceleration of 4.50 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 375 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude (in N) of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 250 kg? Question #: 4.3
Answer: 1500 to 1517.5 (an attempt at an explanation) The method I used got a significant amount of digits lower than the answer and I am not exactly sure why. The way I did this was to first graph the opposing forces, then plug it into F = MA. With this, I got: F - 375 = 4.5(250) ---> F = 1125 + 375 ---> F = 1500 I will be updating this the moment I find a better answer.
A powerful motorcycle can produce an acceleration of 3 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 375 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude (in N) of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 250 kg? Question #: 4.3
Answer: 1125 to 1130 (an attempt at an explanation) The method I used got 5 digits lower than the answer and I am not exactly sure why. The way I did this was to first graph the opposing forces, then plug it into F = MA. With this, I got: F - 375 = 3(250) ---> F = 750 + 375 ---> F = 1125 I will be updating this the moment I find a better answer.
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 12,900 kg. The thrust of its engines is 30,500 N. (Assume that the gravitational acceleration on the Moon is 1.67 m/s2.) (a) Calculate (in m/s2) its magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? No, the thrust of the module's engines is less than its weight on Earth. Yes, the thrust of the module's engines is equal to its weight on Earth. No…
(a) 0.694 (b) No, the thrust of the module's engine is less than its weight on Earth (c) NONE an attempt at an explanation -- For (a) -- First things first, we have all the numbers we need for F = MA, however, the only thing to account for is the opposing force (Weight) on the moon. We'll need to use W = mg in order to add this opposing force. W = 12,900(1.67) --> W = -21543 \\this has to be negative when inputted into the formula because it opposes the positive Y-axis force F +/- W = MA 30,500…
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,300 kg. The thrust of its engines is 33,000 N. (Assume that the gravitational acceleration on the Moon is 1.67 m/s2.) (a) Calculate (in m/s2) its magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? No, the thrust of the module's engines is less than its weight on Earth. Yes, the thrust of the module's engines is equal to its weight on Earth. No…
(a) 1.53 (b) No, the thrust of the module's engines is less than its weight on Earth. (c) NONE (an attempt at an explanation) -- For (a) -- First things first, we have all the numbers we need for F = MA, however, the only thing to account for is the opposing force (Weight) on the moon. We'll need to use W = mg in order to add this opposing force. W = 12,900(1.67) --> W = -21543 \\this has to be negative when inputted into the formula because it opposes the positive Y-axis force F +/- W = MA 30,…
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…
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…
The wheels of a midsize car exert a force of 2000 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 255 N and the acceleration of the car is 1.50 m/s2, what is the mass (in kg) of the car plus its occupants? Question #: 4.8
Answer: 1173.33 (an attempt at an explanation) Overall, this is a super EZ plug-n'-play question. We just plug all of this into F=MA and we'll get the answer! The only thing to keep note is that, even though the force is exerted backward, the 2000N is still giving an opposite and equal reaction -> forcing the car forwards (which thusly means it's positive, not negative!) And since this only worries itself on the X-axis, we don't care about gravity and weight! 2000 - 255 = m(1.5) 1745/1.5 = m m…
The wheels of a midsize car exert a force of 2000 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 270 N and the acceleration of the car is 1.50 m/s2, what is the mass (in kg) of the car plus its occupants? Question #: 4.8
Answer: 1153.33 (an attempt at an explanation) Overall, this is a super EZ plug-n'-play question. We just plug all of this into F=MA and we'll get the answer! The only thing to keep note is that, even though the force is exerted backward, the 2000N is still giving an opposite and equal reaction -> forcing the car forwards (which thusly means it's positive, not negative!) And since this only worries itself on the X-axis, we don't care about gravity and weight! 2000 - 270 = m(1.5) 1730/1.5 = m m…
The wheels of a midsize car exert a force of 2050 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 230 N and the acceleration of the car is 1.50 m/s2, what is the mass (in kg) of the car plus its occupants? Question #: 4.8
Answer: (an attempt at an explanation) Overall, this is a super EZ plug-n'-play question. We just plug all of this into F=MA and we'll get the answer! The only thing to keep note is that, even though the force is exerted backward, the 2000N is still giving an opposite and equal reaction -> forcing the car forwards (which thusly means it's positive, not negative!) And since this only worries itself on the X-axis, we don't care about gravity and weight! 2050 - 230 = m(1.5) 1820/1.5 = m m = 1213.33
THIS WILL NOT BE ON THE EXAM ANYMORE the easiest question will not be on the exam D: To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the road (the rolling resistance). Assume all values are accurate to three significant digits. (Assume the density of air is 1.21 kg/m3.) (a) What is the magnitude of drag force (in N) at 69 km/h and 114 km/h for a Toyota Camry? (Drag area is 0.70 m2, and drag coefficient is 0.28.) This is…
(a) First (Camry) question set For 69 km/h, the total = 43.6N For 114 km/h, total = 119.2N (b) Second (Hummer) Question set For 69 km/h, the total = 347N For 114 km/h, total = 949.4N (an attempt at an explanation) -- For (a) -- This is basically a plug n' play question with only a few unit conversions for Meters/second. We start by using this equation and list out our units which are: Fd = 1/2 C p A V^2 <-- applicable units have to be in m/s C = Drag Coefficient which is 0.28 p = Density of air…
THIS WILL NOT BE ON THE EXAM ANYMORE the easiest question will not be on the exam D: To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the road (the rolling resistance). Assume all values are accurate to three significant digits. (Assume the density of air is 1.21 kg/m3.) (a) What is the magnitude of drag force (in N) at 71 km/h and 104 km/h for a Toyota Camry? (Drag area is 0.70 m2, and drag coefficient is 0.28.) This is…
(a) First (Camry) question set For 69 km/h, the total = 46.1 For 114 km/h, total = 99N (b) Second (Hummer) Question set For 69 km/h, the total = 367N For 114 km/h, total = 788 (an attempt at an explanation) -- For (a) -- This is basically a plug n' play question with only a few unit conversions for Meters/second. We start by using this equation and list out our units which are: Fd = 1/2 C p A V^2 <-- applicable units have to be in m/s C = Drag Coefficient which is 0.28 p = Density of air which…
The propeller of a World War II fighter plane is 2.65 m in diameter. (a) What is its angular velocity in radians per second if it spins at 1500 rev/min? (b) What is the linear speed (in m/s) of its tip at this angular velocity if the plane is stationary on the tarmac? (c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of G. (1c) Find for Centripetal Acceleration in m/s^2 & (2c) Find for Acceler…
(a) 157.1 Rad/s (b) 208.16 (1c) 32,702.33 (2c) 3336.97 (an attempt at an explanation) -- For (a) -- This is actually just a unit-conversion question. All we need to do is get it into radians per second. Thusly, we will do the following: 1500 rev/min * (2π)rad/1rev * 1min/ 60s or (1500 * 2π)/60 = ~157.1 -- For (b) -- We would have to use a certain formula to help us out, that being: linear speed (V) = radius (r) * angular velocity (w) or v = rw after this, it is a super easy plug-n'-play! We kno…
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…
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…
What is the ideal speed to take a 80 m radius curve banked at a 10.0° angle? Question #: 6.8
Answer: 11.76 (an attempt at an explanation) If we know the following formula, this is an EZ question since we have the units we want: tan(theta) = (v^2)/rg Once we put it all in a list, we got this! theta = 10 r = 80 g = 9.8 Since we're finding V, this is easy! tan(10) = (v^2) / 80(9.8) tan(10) = (v^2) / 784 784tan(10) = v^2 root[784tan(10)] = v v = ~11.76
What is the ideal speed to take a 90 m radius curve banked at a 10.0° angle? Question #: 6.8
Answer: 12.5 (an attempt at an explanation) If we know the following formula, this is an EZ question since we have the units we want: tan(theta) = (v^2)/rg Once we put it all in a list, we got this! theta = 10 r = 90 g = 9.8 Since we're finding V, this is easy! tan(10) = (v^2) / 90(9.8) tan(10) = (v^2) / 882 882tan(10) = v^2 root[882tan(10)] = v v = ~12.471 --> 12.5
The propeller of a World War II fighter plane is 2.48 m in diameter. (a) What is its angular velocity in radians per second if it spins at 1500 rev/min? (b) What is the linear speed (in m/s) of its tip at this angular velocity if the plane is stationary on the tarmac? (c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of G. (1c) Find for Centripetal Acceleration in m/s^2 & (2c) Find for Acceler…
(a) 157.1 Rad/s (b) 195 (rounded) (1c) 30,600 (rounded) (2c) 3,120 (rounded) (an attempt at an explanation) -- For (a) -- This is actually just a unit-conversion question. All we need to do is get it into radians per second. Thusly, we will do the following: 1500 rev/min * (2π)rad/1rev * 1min/ 60s or (1500 * 2π)/60 = ~157.1 -- For (b) -- We would have to use a certain formula to help us out, that being: linear speed (V) = radius (r) * angular velocity (w) or v = rw after this, it is a super eas…
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. Work in progress!!!
(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: Vfy^2 = Viy^2 + 2ay (y-yo) 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 = root[52920] Vfy = ~230 m/s --…
Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14.5° slope (one that makes an angle of 14.5° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration. (a) on dry concrete (1) (b) on wet concrete (0.7) (c) on ice (0.1) Question #5.2
(a) 7.03 (b) 4.19 (c) -1.5 ---------- only an answer if you can understand my insanity step 1: make graphic step 2: make graphic + all things we know step 3: add components of W (x and Y) \\WY must be negative. Y? idk, prob cus it semi-opposes forces(?) or it is leading strait to neg Y axis. \\Force must oppose wy upward. |wy| is that force (otherwise car go under grnd) step 4: use sin for wx \\they are opposite cus axes on w are different step 5: input into f=ma \\ we see we need fs (stat fric…
Calculate the maximum acceleration (in m/s2) of a car that is heading up a 4.4° slope (one that makes an angle of 4.4° with the horizontal) under the following road conditions. Assume the full weight of the car is supported by the four drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling. Enter your answers to at least three significant figures.) (a) on dry concrete (1) (b) on wet concrete (0.7)…
(a) 9.019 (b) 6.088 (c) 0.2253 ---------------- only an answer if you can understand my insanity step 1: make graphic step 2: make graphic + all things we know step 3: add components of W (x and Y) \\WY must be negative. Y? idk, prob cus it semi-opposes forces(?) or it is leading strait to neg Y axis. \\Force must oppose wy upward. |wy| is that force (otherwise car go under grnd) step 4: use sin for wx \\they are opposite cus axes on w are different step 5: input into f=ma \\ we see we need fs…