course Phy 121 This should be the remainder of assignment 24. ??????w????assignment #024
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14:25:46 Why was it necessary to let the string go slack at the top of the circle in order to get the desired results?
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RESPONSE --> Slack introduced into the string at the top of its circular path when the weight gets to the top of its path means that we know with a reasonable degree of certainty that the centripetal acceleration is equal to gravity acceleration. Tension in the string causes the weight at the end of the string to be pulled down and toward the center.
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14:26:01 ** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity. If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight. If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. **
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RESPONSE --> My answer is at least partially correct.
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14:30:06 Why do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal?
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RESPONSE --> When a ball is attached to a string moving in a circular fashion similar to that previously mentioned, at any point in the circle the instantaneous velocity is in a direction tangent to the circular path.
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14:31:46 ** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity, being perpendicular to this vertical, must be entirely in the horizontal direction. **
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RESPONSE --> I failed to mention the radial line, but the concept of the tangent line does sort of relate here.
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14:35:43 What is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force?
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RESPONSE --> Ar = v^2 / r The source of the force is centripetal acceleration.
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14:36:11 ** Under these conditions, with the string slack and not exerting any force on the object, the centripetal acceleration will be equal to the acceleration of gravity. **
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RESPONSE -->
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14:43:56 Query principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin?
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RESPONSE --> If you diagram this out it looks sort of like a square. If you draw a point of origin and a line in the positive y direction which represents 18 blocks north, and then another line in the positive x direction that represents 10 blocks east, and then another line in the negative y direction which represents 16 blocks south. I'm going to say displacement is 12 blocks.
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14:47:25 The final position of the truck is 2 blocks south and 10 blocks east. This is equivalent to a displacement of +10 blocks in the x direction and -2 blocks in the y direction. The distance is therefore sqrt( (10 blocks)^2 + (-2 blocks)^2 ) = sqrt( 100 blocks^2 + 4 blocks^2) = sqrt(104 blocks^2) = sqrt(104) * sqrt(blocks^2) = 10.2 blocks. The direction makes and angle of theta = arcTan(-2 blocks / (10 blocks) ) = arcTan(-.2) = -12 degrees with the positive x axis, as measured counterclockwise from that axis. This puts the displacement at an angle of 12 degrees in the clockwise direction from that axis, or 12 degrees south of east.
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RESPONSE --> Okay, my guess was pretty close to what the answer actually was, however I don't understand why 10 and 2 are squared and added together and then you take the square root of that answer. Why? I guess what I really don't understand is how you know to perform the mathematical calculations presented in the solution. Is there some sort of rule that I'm missing here?
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14:54:02 Query principles of physics and general college physics 7.18: Diver leaves cliff traveling in the horizontal direction at 1.8 m/s, hits the water 3.0 sec later. How high is the cliff and how far from the base does he land?
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RESPONSE --> He lands 5.4 meters from the base. I know this because at a horizontal direction of 1.8m/s after three seconds the distance (1.8*3) is 5.4m. The height of the cliff was determined using a^2 + b^2 = c^2. 5.4^2 + 1.8^2 = c^2 32.4 = c^2 c = 5.7 m
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14:56:33 The diver's initial vertical velocity is zero, since the initial velocity is horizontal. Vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. We will choose downward as the positive direction, so the vertical motion has v0 = 0, constant acceleration 9.8 m/s^2 and time interval `dt = 3.0 seconds. The third equation of uniformly accelerated motion tells us that the vertical displacement is therefore vertical motion: `ds = v0 `dt + .5 a `dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m. The cliff is therefore 44 m high. The horizontal motion is characterized 0 net force in this direction, resulting in horizontal acceleration zero. This results in uniform horizontal velocity so in the horizontal direction v0 = vf = vAve. Since v0 = 1.8 m/s, vAve = 1.8 m/s and we have horizontal motion: `ds = vAve * `dt = 1.8 m/s * 3.0 s = 5.4 meters.
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RESPONSE --> I got the distance from the base and missed the height of the cliff. I forgot to take gravity into consideration, and I didn't use the formula: 'ds = v0 'dt + .5 a 'dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m
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14:56:52 Gen phy 3.13 A 44 N at 28 deg, B 26.5 N at 56 deg, C 31.0 N at 270 deg. Give your solution to the problem.
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RESPONSE --> Not applicable; I am a principles of physics student.
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14:57:02 ** The solution given here is for a previous edition, in which the forces are Force A of 66 at 28 deg Force B of B 40 at 56 deg Force C of 46.8 at 270 deg These forces are very close to 2/3 as great as the forces given in the current edition, and all correct results will therefore be very close to 2/3 as great as those given here. Calculations to the nearest whole number: A has x and y components Ax = 66 cos(28 deg) = 58 and Ay = 66 sin(28 deg) = 31 Bhas x and y components Bx = 40 cos(124 deg) = -22 and By = 40 sin(124 deg) = 33 C has x and y components Cx = 46.8 cos(270 deg) = 0 and Cy = 46.8 sin(270 deg) = -47 A - B + C therefore has components Rx = Ax-Bx+Cx = 58 - (-22) + 0 = 80 and Ry = Ay - By + Cy = 31-33-47=-49, which places it is the fourth quadrant and gives it magnitude `sqrt(Rx^2 + Ry^2) = `sqrt(80^2 + (-49)^2) = 94 at angle tan^-1(Ry / Rx) = tan^-1(-49/53) = -32 deg or 360 deg - 32 deg = 328 deg. Thus A - B + C has magnitude 93 at angle 328 deg. B-2A has components Rx = Bx - 2 Ax = -22 - 2 ( 58 ) = -139 and Ry = By - 2 Ay = 33 - 2(31) = -29, placing the resultant in the third quadrant and giving it magnitude `sqrt( (-139)^2 + (-29)^2 ) = 142 at angle tan^-1(Ry / Rx) or tan^-1(Ry / Rx) + 180 deg. Since x < 0 this gives us angle tan^-1(-29 / -139) + 180 deg = 11 deg + 180 deg = 191 deg. Thus B - 2 A has magnitude 142 at angle 191 deg. Note that the 180 deg is added because the angle is in the third quadrant and the inverse tangent gives angles only in the first or fourth quandrant ( when the x coordinate is negative we'll be in the second or third quadrant and must add 180 deg). **
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RESPONSE --> Not applicable.
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14:57:10 Univ. 3.58. (This problem has apparently been eliminated from recent editions, due to the now policitally incorrect nature of the device being thrown. The problem is a very good one and has been edited to eliminate politically incorrect references). Good guys in a car at 90 km/hr are following
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RESPONSE --> Not applicable.
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15:08:23 bad guys driving their car, which at a certain instant is 15.8 m in front of them and moving at a constant 110 km/hr; an electronic jamming device is thrown by the good guys at 45 deg above horizontal, as they observe it. This device must land in the bad guy's car. With what speed must the device be thrown relative to the good guys, and with what speed relative to the ground?
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RESPONSE --> I really don't know how to begin an approach to answer this question correctly.
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15:10:16 ** The device is thrown at velocity v0 at 45 deg, giving it v0y = .71 v0 and v0x = .71 v0. The device will return to its original vertical position so we have `dsy = 0. Using `dsy = v0y `dt + .5 g `dt^2 with `dsy = 0 and assuming the upward direction to be positive we obtain v0y `dt + .5 (-g) `dt^2 = 0 so that `dt = 0 or `dt = - 2 * v0y / (-g) = 2 * 71 v0 / g. In time `dt the horizontal displacement relative to the car will be `dsx = v0x `dt + ax `dt; since acceleration ax in the x direction and v0x = .71 v0 is zero we have `dsx = .71 v0 * `dt. We also know that relative to the first car the second is moving at 20 km / hr = 20,000 m / (3600 sec) = 5.55 m/s, approx.; since its initial position is 15.8 m in front of the first car we have `dsx = 15.8 m + 5.55 m/s * `dt. To keep the equations symbolic we use x0Relative and vRelative for the relative initial position and velocity of the second car with respect to the first. We thus have three equations: `dt = 2 * .71 v0 / g = 1.42 v0 / g. `dsx = .71 v0 * `dt `dsx = x0Relative + v0Relative * `dt. This gives us three equations in the variables v0, `dt and `dsx, which we reduce to two by substituting the expression -2 to obtain: `dsx = .71 v0 * 1.42 v0 / g = v0^2 / g `dsx = x0Relative + v0Relative * 1.42 v0 / g. Setting the right-hand sides equal we have v0^2 / g = x0Relative + v0Relative * 1.42 v0 / g, or v0^2 - v0Relative * 1.42 v0 - g * x0Relative = 0. We get v0 = [1.42 v0Relative +-sqrt( (1.42 v0Relative)^2 - 4 * (-g * x0Relative) ) ] / 2 = [1.42 * v0Relative +-sqrt( (1.42 * v0Relative)^2 + 4 * g * x0Relative) ] / 2. Substituting 5.55 m/s for v0Relative and 15.8 m for x0Relative we get [1.42 * 5.55 m/s +-sqrt( (1.42 *5.55 m/s)^2 + 4 * 9.8 m/s^2 *15.8 m) ] / 2 = 17 m/s or -9.1 m/s, approx.. We conclude that the initial velocity with respect to the first case must be 17 m/s. Checking this we see that the device will have initial x and y velocities 7.1 * 17 m/s = 12 m/s, approx., and will therefore stay aloft for 2 * 12 m/s / (9.8 m/s^2) = 2.4 sec, approx.. It will therefore travel 2.4 sec * 12 m/s = 28 m, approx. in the horizontal direction relative to the first car. During this time the second car will travel about 5.55 m/s * 2.4 sec = 13 m, approx., resulting in relative position 15.8 m + 13 m = 28.8 m with respect to the first. This is reasonably close to the 28 m obtained from the motion of the projectile. Correcting for roundoff errors will result in precise agreement. **
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RESPONSE --> The use of formulas makes sense, exactly what section was all of this stuff in again?
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