course Phy 121 I would like to know how to solve #1 of Chapter 3. The direction Southwest confused me. I'm used to working with a vertical direction and a horizontal direction. I didn't really know what to do.
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17:16:42 `q001. Note that this assignment contains 4 questions. . Note that this assignment contains 4 questions. When an object moves a constant speed around a circle a force is necessary to keep changing its direction of motion. This is because any change in the direction of motion entails a change in the velocity of the object. This is because velocity is a vector quantity, and if the direction of a vector changes, then the vector and hence the velocity has changed. The acceleration of an object moving with constant speed v around a circle of radius r has magnitude v^2 / r, and the acceleration is directed toward the center of the circle. This results from a force directed toward the center of the circle. Such a force is called a centripetal (meaning toward the center) force, and the acceleration is called a centripetal acceleration. If a 12 kg mass travels at three meters/second around a circle of radius five meters, what centripetal force is required?
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RESPONSE --> centripetal acceleration = v^2/r m = 12kg v = 3m/s r = 5m First we find the centripetal acceleration with the formula v^2/r. a = (3m/s)^2 / 5m = 1.8m/s^2 Now that we know the acceleration and the mass, we can find the centripetal force. F = m*a = 12kg * 1.8m/s^2 = 21.6 Newtons
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17:16:57 The centripetal acceleration of the object is v^2 / r = (3 meters/second) ^ 2/(5 meters) = 1.8 meters/second ^ 2. The centripetal force, by Newton's Second Law, must therefore be Fcent = 12 kg * 1.8 meters/second ^ 2.
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RESPONSE --> I understand.
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17:27:44 `q002. How fast must a 50 g mass at the end of a string of length 70 cm be spun in a circular path in order to break the string, which has a breaking strength of 25 Newtons?
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RESPONSE --> m = 50g = 0.05kg r = 70cm = 0.7m F = 25N We can use the formula that was just introduced for centripetal accleration to find the velocity of the object. a = v^2/r F = m*a = m(v^2/r) 25N = (0.05kg)(v^2/0.7m) 500N/kg = v^2/0.7m v^2 = 350m^2/s^2 v = 18.71 m/s
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17:28:04 The centripetal acceleration as speed v will be v^2 / r, where r = 70 cm = .7 meters. The centripetal force will therefore be m v^2 / r, where m is the 50 g = .05 kg mass. If F stands for the 25 Newton breaking force, then we have m v^2 / r = F, which we solve for v to obtain v = `sqrt(F * r / m). Substituting the given values we obtain v = `sqrt( 25 N * .7 meters / (.05 kg) ) = `sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) = `sqrt(350 m^2 / s^2) = 18.7 m/s.
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RESPONSE --> Ok.
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17:33:08 `q003. What is the maximum number of times per second the mass in the preceding problem can travel around its circular path before the string breaks?
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RESPONSE --> We know that the breaking force is 25 Newtons. In order to know how many times the mass revolves, we must find the diameter of the circular path. we know that the radius is 0.7m. To find diameter we use the formula d = 2pi*r. d = 2pi*0.7m = 4.398 m Now we know the distance the object travels in one revolution. We can use this distance and the velocity at which it travels, 18.7m/s, to find the number of times it revolves. `ds = vAve * `dt `dt = `ds/vAve = (18.7m/s)/(4.398m) = 4.25 revolutions per second.
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17:33:21 The maximum possible speed of the mass was found in the preceding problem to be 18.7 meters/second. The path of the mass is a circle of radius 70 cm = .7 meters. The distance traveled along this path in a single revolution is 2 `pi r = 2 `pi * .7 meters = 4.4 meters, approximately. At 18.7 meters/second, the mass will travel around the circle 18.7/4.4 = 4.25 times every second.
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RESPONSE --> I understand this problem.
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17:34:17 `q004. Explain in terms of basic intuition why a force is required to keep a mass traveling any circular path.
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RESPONSE --> As the mass changes direction as it does constantly in a circular path, the velocity changes. In order to keep the object moving and changing direction, a constant force must be applied to it.
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17:34:36 We simply can't change the direction of motion of a massive object without giving it some sort of a push. Without such a force an object in motion will remain in motion along a straight line and with no change in speed. If your car coasts in a circular path, friction between the tires and the road surface pushes the car toward the center of the circle, allowing it to maintain its circular path. If you try to go too fast, friction won't be strong enough to keep you in the circular path and you will skid out of the circle. In order to maintain a circular orbit around the Earth, a satellite requires the force of gravity to keep pulling it toward the center of the circle. The satellite must travel at a speed v such that v^2 / r is equal to the acceleration provided by Earth's gravitational field.
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RESPONSE --> I understand this concept.
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v????????e?q? assignment #024 024. `query 24 Physics I 07-11-2006
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19:24:04 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 --> I'm not sure I know what string or circle this question is asking about. The experiment for this assignment was the Conservation of Momentum. So I am assuming this question is just asking about any situation like this. I would think that the string would have to be slack in order to avoid any variation of the centripetal acceleration. If there is no tension on the string, the centripetal acceleration would be equal to the gravitational acceleration because it would move just like a free falling object.
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19:26:23 ** 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 --> Ok.
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19:28:41 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 --> The initial velocity would be horizontal at this point because if the washer begins at the top of the circle it will move horizontally to the center of the circle.
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19:31:37 ** 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 --> Ok. I don't really understand this concept. Does the radial line represent anything?
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19:33:31 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 --> The centripetal acceleration at this point would be equal to the acceleration of gravity which is 9.8m/s^2. The source of this force, therefore, is gravity itself.
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19:33:47 ** 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 --> Ok
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19:35:49 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 --> Drawing a diagram of this problem helped me to understand. If the truck drives 18 blocks north and then later 16 blocks south, it will be end up 2 blocks north in the vertical direction. The horizontal direction is only 10 blocks east. So we can think of these directions as the legs of a right triangle. We solve using the Pythagorean Theorem: c^2 = (2)^2 + (10)^2 = 104 c = 10.20 blocks
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19:37:18 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 --> I understand this problem. I don't think the question asked for the direction, so I did not calculate it. But I do understand how to calculate the angle of direction. Also, why is the vertical direction in the solution negative. I thought it was positive because the north displacement is greater than the south displacement.
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19:40:03 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 --> horizontal v = 1.8m/s vo = 0m/s (free falling) `dt = 3.0s g = 9.8m/s^2 Height of cliff = vertical `ds: `ds = (v0)(`dt) + 1/2(g)(`dt^2) `ds = (0m/s)(3s) + 1/2(9.8m/s^2)(3s)^2 = 44.1 meters Distance from base = horizontal `ds: `ds = vAve * `dt `ds = (1.8m/s)(3s) = 5.4 meters
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19:40:53 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 understand.
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19:41:42 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 --> This text problem was not assigned to my course, Physics 121.
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19:41:50 ** 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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19:42:17 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 --> This text problem was not assigned to me. I am a Principles of Physics student.
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19:42:26 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 -->
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19:42:30 ** 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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