course Phy231 ?z?????I????w?_??assignment #006006. Using equations with uniformly accelerated motion.
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18:10:39 `q001. Note that there are 9 questions in this assignment. Using the equation vf = v0 + a * `dt determine the acceleration of an object whose velocity increases at a uniform rate from 10 m/s to 30 m/s in 15 seconds. Begin by solving the equation for the acceleration a, then 'plug in' your initial and final velocities. Describe your work step y step.
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RESPONSE --> vf=v0 + a * `dt a*`dt=vf-v0 a=(vf-v0)/`dt a=(30m/s-10m/s)/15s=1.33m/s^2 confidence assessment: 3
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18:10:46 The equation vf = v0 + a * `dt is solved for a by first adding -v0 to both sides to obtain vf - v0 = v0 + a * `dt - v0, which simplifies to vf - v0 = a * `dt. Both sides are then divided by `dt to obtain (vf - v0) / `dt = a. Reversing left-and right-hand sides we obtain the formula a = (vf - v0) / `dt. We then plug in our given values of initial and final velocities and the time interval. Since velocity increases from 10 m/s to 30 m/s, initial velocity is v0 = 10 m/s and final velocity is vf = 30 m/s. The time interval `dt is 15 seconds, so we have a = (30 m/s - 10 m/s) / (15 s) = 20 m/s / (15 s) = 1.33.. m/s^2.
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RESPONSE --> self critique assessment: 3
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18:11:14 `q002. It wasn't necessary to use a equation to solve this problem. How could this problem had been reasoned out without the use of an equation?
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RESPONSE --> Calculate the change in velocity then divide that by the change in time confidence assessment: 3
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18:11:39 Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second. We would then divided change in velocity by the time interval to get 20 meters/second / (15 sec) = 1.33 m/s^2.
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RESPONSE --> self critique assessment: 3
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18:14:34 `q003. Use the equation `ds = (vf + v0) / 2 * `dt to determine the initial velocity of an object which accelerates uniformly through a distance of 80 meters in 10 seconds, ending up at a velocity of 6 meters / sec. begin by solving the equation for the desired quantity. Show every step of your solution.
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RESPONSE --> `ds=(vf+v0)/2*`dt `ds/`dt=(vf+v0)/2 2*`ds/`dt=vf+v0 v0=2*`ds/`dt-vf v0=2*80m/10s-6m/s v0=16m/s-6m/s=10m/s confidence assessment: 3
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18:14:41 We begin by solving the equation for v0. Starting with `ds = (vf + v0) / 2 * `dt, we can first multiply both sides of the equation by 2 / `dt, which gives us `ds * 2 / `dt = (vf + v0) / 2 * `dt * 2 / `dt. The right-hand side can be rearranged to give (vf + v0) * `dt / `dt * 2 / 2; since `dt / `dt = 1 and 2 / 2 = 1 the right-hand side becomes just vf + v0. The equation therefore becomes 2 * `ds / `dt = vf + v0. Adding -vf to both sides we obtain v0 = 2 * `ds / `dt - vf. We now plug in `ds = 80 meters, `dt = 10 sec and vf = 6 m/s to get v0 = 2 * 80 meters / 10 sec - 6 m/s = 160 meters / 10 sec - 6 m/s = 16 m/s - 6 m/s = 10 m/s.
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RESPONSE --> self critique assessment: 3
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18:20:31 `q004. We can reconcile the above solution with straightforward reasoning. How could the initial velocity have been reasoned out from the given information without the use of an equation? Hint: two of the quantities given in the problem can be combined to give another important quantity, which can then be combined with the third given quantity to reason out the final velocity.
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RESPONSE --> The starting velocity is equal to the final velocity minus the change in the velocity. The change in velocity is equal to the change in displacement divided by the change in time. confidence assessment: 3
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18:22:52 The average velocity of the object is the average rate at which its position changes, which is equal to the 80 meters change in position divided by the 10 s change in clock time, or 80 meters / 10 sec = 8 meters / sec. Since the 8 m/s average velocity is equal to the average of the unknown initial velocity and the 6 m/s final velocity, we ask what quantity when average with 6 m/s will give us 8 m/s. Knowing that the average must be halfway between the two numbers being averaged, we see that the initial velocity must be 10 m/s. That is, 8 m/s is halfway between 6 m/s and 10 m/s.
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RESPONSE --> self critique assessment: 3
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18:26:36 `q005. Using the equation `ds = v0 `dt + .5 a `dt^2 determine the initial velocity of an object which accelerates uniformly at -2 m/s^2, starting at some unknown velocity, and is displaced 80 meters in 10 seconds. Begin by solving the equation for the unknown quantity and show every step.
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RESPONSE --> `ds=v0(`dt)+.5(a)(`dt^2) `ds-.5(a)(`dt^2)=v0*`dt v0=[`ds-.5(a)(`dt^2)]/`dt v0=[80m-.5(-2m/s^2)(10s^2)]/10s v0=[80m+100m]/10s v0=180m/10s=18m/s confidence assessment: 3
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18:26:40 The unknown quantity is the initial velocity v0. To solve for v0 we start with `ds = v0 `dt + .5 a `dt^2. We first add -.5 a `dt^2 to both sides to obtain `ds - .5 a `dt^2 = v0 `dt. We then divide both sides by `dt to obtain (`ds - .5 a `dt^2) / `dt = v0. Then we substitute the given displacement `ds = 80 meters, acceleration a = -2 m/s^2 and time interval `dt = 10 seconds to obtain v0 = [ 80 meters - .5 * (-2 m/s^2) * (10 sec)^2 ] / (10 sec) = [ 80 meters - .5 * (-2 m/s^2) * 100 s^2 ] / (10 sec) = [ 80 m - (-100 m) ] / (10 sec) = 180 m / (10 s) = 18 m/s.
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RESPONSE --> self critique assessment: 3
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18:28:23 `q006. Check the consistency of this result by verifying, by direct reasoning rather than equations, that an object whose initial velocity is 18 m/s and which accelerates for 10 seconds at an acceleration of -2 m/s^2 does indeed experience a displacement of 80 meters.
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RESPONSE --> This doesn't match the previous question. The previous question gave a, `ds and `dt and asked to determine v0 confidence assessment: 0
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18:28:36 The change in the velocity of the object will be -2 m/s^2 * 10 s = -20 m/s. The object will therefore have a final velocity of 18 m/s - 20 m/s = -2 m/s. Its average velocity will be the average (18 m/s + (-2 m/s) ) / 2 = 8 m/s. An object which travels at an average velocity of 8 m/s for 10 sec will travel 80 meters.
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RESPONSE --> self critique assessment: 3
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12:35:08 `q007. Using the equation vf^2 = v0^2 + 2 a `ds determine the initial velocity of an object which attains a final velocity of 20 meters/second after accelerating uniformly at 2 meters/second^2 through a displacement of 80 meters. Begin by solving the equation for the unknown quantity and show every step.
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RESPONSE --> vf^2=v0^2+2a`ds v0^2=vf^2-2a`ds v0=sqrt(vf^2-2a`ds) v0=sqrt((20m/s)^2-2(2m/s^2)(80m)) v0=sqrt(400m^2/s^2-320m^2/s^2) v0=sqrt(80m^2/s^2) v0=8.9m/s confidence assessment: 3
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12:35:13 To solve for the unknown initial velocity v0 we start with vf^2 = v0^2 + 2 a `ds. We first add -2 a `ds to both sides to obtain vf^2 - 2 a `ds = v0^2. We then reverse the right-and left-hand sides and take the square root of both sides, obtaining v0 = +- `sqrt( vf^2 - 2 a `ds). We then substitute the given quantities vf = 20 m/s, `ds = 80 m and a = 3 m/s^2 to obtain v0 = +- `sqrt( (20 m/s)^2 - 2 * 2 m/s^2 * 80 m) = +- `sqrt( 400 m^2 / s^2 - 320 m^2 / s^2) = +- `sqrt(80 m^2 / s^2) = +- 8.9 m/s (approx.).
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RESPONSE --> self critique assessment: 3
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12:38:21 `q008. We can verify that starting at +8.9 m/s an object which attains a final velocity of 20 m/s while displacing 80 meters must accelerate at 2 m/s^2. In this case the average velocity will be ( 8.9 m/s + 20 m/s) / 2 = 14.5 m/s (approx) and the change in velocity will be 20 m/s - 8.9 m/s = 11.1 m/s. At average velocity 14.5 meters/second the time required to displace the 80 meters will be 80 m / (14.5 sec) = 5.5 sec (approx). The velocity change of 11.1 meters/second in 5.5 sec implies an average acceleration of 11.1 m/s / (5.5 sec) = 2 m/s^2 (approx), consistent with our results. Verify that starting at -8.9 m/s the object will also have acceleration 2 meters/second^2 if it ends up at velocity 20 m/s while displacing 80 meters.
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RESPONSE --> vf^2=v0^2+2a`ds (20m/s)^2=(-8.9m/s)^2+2(2m/s^2)(80m) 400m^2/s^2=79.21m^2/s^2+320m^2/s^2 400m^2/s^2=399.21m^2/s^2(approx) confidence assessment: 3
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12:38:26 In this case the average velocity will be ( -8.9 m/s + 20 m/s) / 2 = 5.5 m/s (approx) and the change in velocity will be 20 m/s - (-8.9 m/s) = 28.9 m/s (approx). At average velocity 5.5 meters/second the time required to displace the 80 meters will be 80 m / (5.5 sec) = 14.5 sec (approx). The velocity change of 28.5 meters/second in 14.5 sec implies an average acceleration of 28.5 m/s / (14.5 sec) = 2 m/s^2 (approx), again consistent with our results.
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RESPONSE --> self critique assessment: 3
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12:39:43 `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does?
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RESPONSE --> Initially the object is traveling south at 8.9m/s. As time passes it is accelerating northward at 2m/s^2. Eventually this overcomes the negative southward velocity and the object heads north. confidence assessment: 3
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12:39:47 The object ends up at position +80 meters, which is assumed to be 80 meters to the North. Its initial velocity is -8.9 m/s, with the - sign indicating that the initial velocity is in the direction opposite to the displacement of the object. So the object must start out moving to the South at 8.9 meters/second. Its acceleration is +2 m/s^2, which is in the opposite direction to its initial velocity. This means that the velocity of the object changes by +2 m/s every second. After 1 second the velocity of the object will therefore be -8.9 m/s + 2 m/s = -6.9 m/s. After another second the velocity will be -6.9 m/s + 2 m/s = -4.9 m/s. After another second the velocity will be -2.9 m/s, after another -.9 m/s, and after another -.9 m/s + 2 m/s = +1.1 m/s. The speed of the object must therefore decrease, starting at 8.9 m/s (remember speed is always positive because speed doesn't have direction) and decreasing to 6.9 m/s, then 4.9 m/s, etc. until it reaches 0 for an instant, and then starts increasing again. Since velocities after that instant become positive, the object will therefore start moving to the North immediately after coming to a stop, picking up speed at 2 m/s every second. This will continue until the object has attained a velocity of +20 meters/second and has displaced +80 meters from its initial position.{}{}It is important to understand that it is possible for velocity to be in one direction and acceleration in the other. In this case the initial velocity is negative while the acceleration is positive. If this continues long enough the velocity will reach zero, then will become positive.
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RESPONSE --> self critique assessment: 3
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course Phy231 ?v???????\???assignment #006006. `query 6
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15:29:21 General Physics 1.42. At 1200 liters/day per family how much would level of 50 km^2 lake fall in a year if supplying town of population 40000
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RESPONSE --> Depends on how deep the lake is. Town consumption=1200l/d/fam*(40000pop/3pop/fam)=1.6x10^7 l/day * 365days/year=5.84x10^9 l/year 5.84x10^9 liters * .001 m^3/1 liter=5.84x10^7 m^3 50 km^2 * (1000m)^2/km^2 = 5.0x10^7 m^2 5.84x10^7 m^3 / 5.0x10^7 m^2= 1.168m confidence assessment: 3
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15:30:25 ** A liter is 1/1000 of a cubic meter. It can be thought of as a cube 10 cm on a side. To fill a 1-meter cube t would take 10 rows with 10 cubes in each row to make a single layer 10 cm high, and 10 layers to fill the cube--i.e., 10 * 10 * 10 = 1000 one-liter cubes fill a 1-meter cube. A km is 1000 meters so a km^2 is (1000 m)^2 = 10^6 m^2. 1200 liters / day per family, with 40,000 people at 4 persons / family, implies 10,000 families each using 1200 * .001 m^3 = 1.2 m^3 per day. Total usage would be 10,000 families * 1.2 m^3 / day / family * 365 days / year = 4.3 * 10^6 m^3 / year. The volume of the lake corresponding to a depth change `dy is `dy * A, where A is the area of the lake. The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2. `dy * A = Volume so `dy = Volume / A = 4.3 * 10^6 m^3 / (5 * 10^7 m&2) = .086 m or 8.6 cm. This estimate is based on 4 people per family. A different assumption would change this estimate. STUDENT QUESTION: 40000 people in town divided by average household of 4 = 10,000 families 10,000 families * 1200 liters /day = 12000000 liters used per day * 365 days in a year = 4380000000 liter used. Here is where I get confused changing from liters to level of 50 km^2 to subtract. INSTRUCTOR RESPONSE: If you multiply the area of the lake by the change in depth you get the volume of water used. You know the area of the lake and the volume of the water used, from which you can find the change in depth. Of course you need to do the appropriate conversions of units. Remember that a liter is the volume of a cube 10 cm on a side, so it would take 10 rows of 10 such cubes to make one layer, then 10 layers, to fill a cube 1 meter = 100 cm on a side. You should also see that a km^2 could be a square 1 km on a side, which would be 1000 meters on a side, to cover which would require 1000 rows of 1000 1-meters squares. If you end of having trouble with the units or anything else please ask some specific questions and I will try to help you clarify the situation. ANOTHER INSTRUCTOR COMMENT: The water used can be thought of as having been spread out in a thin layer on top of that lake. That thin layer forms a cylinder whose cross-section is the surface of the lake and whose altitude is the change in the water level. The volume of that cylinder is equal to the volume of the water used by the family in a year. See if you can solve the problem from this model. COMMON ERROR: Area is 50 km^2 * 1000 m^2 / km^2 INSTRUCTOR COMMENT: Two things to remember: You can't cover a 1000 m x 1000 m square with 1000 1-meter squares, which would only be enough to make 1 row of 1000 squares, not 1000 rows of 1000 squares. 1 km^2 = 1000 m * 1000 m = 1,000,000 m^2. **
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RESPONSE --> I used different family size estimate. self critique assessment: 3
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15:34:54 univ 1.70 univ sailor 2 km east, 3.5 km SE, then unknown, ends up 5.8 km east find magnitude and direction of 3d leg, explain how diagram shows qualitative agreement
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RESPONSE --> I'm not sure what you are asking for with the term ""qualitative agreement"". The third leg is the only unknown in this problem and can be solved for by since the final vector is known. confidence assessment: 3
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15:35:18 ** Letting the three vectors be A, B and C (C unknown) and the x axis point east we have A at 0 degrees, B at 315 degrees and the resultant at 0 degrees. Ax = 2, Ay = 0 (A is toward the East, along the x axis). Bx = B cos(315 deg) = 2.47, By = B sin(315 deg) = -2.47. Rx = 5.8, Ry = 0. Since Ax + Bx + Cx = Rx we have Cx = Rx - Ax - Bx = 5.8 - 2 - 2.47 = 1.33. Since Ay + By + Cy = Ry we have Cy = Ry - Ay - By = 0 - 0 - (-2.47) = 2.47. C has magnitude sqrt( Cx^2 + Cy^2) = sqrt ( 1.33^2 + 2.47^2) = sqrt(7.9) = 2.8, representing 2.8 km. C has angle arctan(Cy / Cx) = arctan(2.47 / 1.33) = 61 deg. **
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RESPONSE --> Previous problem didn't ask for solution. self critique assessment: 1
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16:02:59 **** query univ 1.82 (1.66 10th edition) vectors 3.6 at 70 deg, 2.4 at 210 deg find scalar and vector product
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RESPONSE --> A=3.6m 70deg A=(cos(70)*3.6,sin(70)*3.6) A=(1.23m,3.38m) B=2.4m 210deg B=(cos(210)*2.4,sin(210)*2.4)) B=(-2.08m,-1.2m) Scalar product= =1.23m*-2.08m+3.38m*-1.2m =-2.56m^2+-4.06m^2 =-6.62m^2 Vector product is not defined for 2D coordinates. Assuming Z values of 0: { (3.38m)(0m)-(0m)(-1.2m) (0m)(-2.08m)-(1.23m)(0m) (1.23m)(-1.2m)-(3.38m)(-2.08m) } = (0m,0m,5.5m) confidence assessment: 3
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16:04:09 ** GOOD STUDENT SOLUTION WITH INSTRUCTOR COMMENT: A * B = |A||B| cos(theta) = AxBx + Ay By + AzBz 3.6 * 2.4 * cos (140 deg) = -6.62 To check for consistency we can calculate the components of A and B: Ax = 3.6 * cos(70 deg) = 1.23 Ay = 3.6 * sin(70 deg) = 3.38 Bx = 2.4 * cos (210 deg) = -2.08 By = 2.4 * sin(210 deg) = -1.2 dot product = 1.23 *-2.08 + 3.38 * -1.2 = -2.5584 + -4.056 = -6.61. Close enough. ---------------------- Cross product: | A X B | = |A| |B| sin (theta) = 3.6 * 2.4 * sin(140) = 5.554. Finding the components we have (Ay * Bz - Az * By) i + (Az * Bx - Ax + Bz) j + (Ax * By - Ay * Bx) k = ((3.38*0)-(0*-1.2)] i + [(0 * -2.08) - (1.23 * 0)] j + [(1.23*-1.2)-(3.38 * -2.08)] k = 0 i + 0 j + 5.55 k, or just 5.55 k, along the positive z axis ('upward' from the plane). INSTRUCTOR COMMENT: Your conclusion is correct. The cross product must be at a right angle to the two vectors. To tell whether the vector is 'up' or 'down' you can use your result, 5.55 k, to see that it's a positive multiple of k and therefore upward. The other way is to use the right-hand rule. You place the fingers of your right hand along the first vector and position your hand so that it would 'turn' the first vector in the direction of the second. That will have your thumb pointing upward, in agreement with your calculated result, which showed the cross product as being in the upward direction. **
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RESPONSE --> self critique assessment: 3
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course Phy231 ??????L??????assignment #007
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16:34:30 `q001. We obtain an estimate of the acceleration of gravity by determining the slope of an acceleration vs. ramp slope graph for an object gliding down an incline. Sample data for an object gliding down a 50-cm incline indicate that the object glides down the incline in 5 seconds when the raised end of the incline is .5 cm higher than the lower end; the time required from rest is 3 seconds when the raised end is 1 cm higher than the lower end; and the time from rest is 2 seconds when the raised end is 1.5 cm higher than the lower end. What is the acceleration for each trial?
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RESPONSE --> 1st trial: vAve=50cm/5s=10cm/s vf=20cm/s a=20cm/s/5s=4cm/s^2 2nd trial: vAve=50cm/3s=16.67cm/s vf=33.34cm/s a=33.34cm/s/3s=11.11cm/s^2 3rd trial: vAve=50cm/2s=25cm/s vf=50cm/s a=50cm/s/2s=25cm/s^2 confidence assessment: 3
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16:34:47 We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.
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RESPONSE --> self critique assessment: 3
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16:35:20 `q002. What are the ramp slopes associated with these accelerations?
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RESPONSE --> s1=0.01 s2=0.02 s3=0.03 confidence assessment: 3
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16:35:25 For the 5-second trial, where acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01. For the 3-second trial, where acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02. For the 2-second trial, where the acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03.
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RESPONSE --> self critique assessment: 3
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16:40:59 `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the straight line that comes as close as possible, on the average, to the three points.
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RESPONSE --> Fitted line has steep slope, meaning as slope increases acceleration increases. The first point seems to be slightly high, the 2nd slightly low and the 3rd moderately higher. confidence assessment: 3
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16:41:25 The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. Or the graph could indicate acceleration vs. ramp slope behavior that is increasing at an increasing rate.
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RESPONSE --> self critique assessment: 3
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16:45:42 `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line?
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RESPONSE --> 0,-13 5,45 =(45-(-13))/(5-0) =58/5=11.6 confidence assessment: 3
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16:46:38 A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way. For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity.
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RESPONSE --> I miscounted my lower point, should have been 0,-6 self critique assessment: 3
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16:23:29 `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last?
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RESPONSE --> 35 seconds cycle=35s/100=0.35s confidence assessment: 3
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16:32:57 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected.
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RESPONSE --> Redid the experiment and got 110 seconds. self critique assessment: 3
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17:09:54 `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity.
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RESPONSE --> T=2*pi*sqrt(L)/sqrt(g) T*sqrt(g)=2*pi*sqrt(L) sqrt(g)=[2*pi*sqrt(L)]/T g=[2*pi*sqrt(L)]^2/T^2 g=[4*pi^2*L]/T^2 g=[4*(3.14)^2*(.3m)]/(1.10s)^2 g=9.78m/s^2 confidence assessment: 3
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17:10:24 Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain {}g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. You should check these calculations for accuracy, since they were mentally approximated.
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RESPONSE --> I used meters. self critique assessment: 2
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