course Phy 201 ????N????????assignment #018018. Vectors
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18:01:04 `q001. Note that this assignment contains 7 questions. . The Pythagorean Theorem: the hypotenuse of a right triangle has a length c such that c^2 = a^2 + b^2, where a and b are the lengths of the legs of the triangle. Sketch a right triangle on a set of coordinate axes by first locating the point (7, 13). Then sketch a straight line from the origin of the coordinate system to this point to form the hypotenuse of the triangle. Continue by sketching a line straight down from (7, 13) to the x axis to form one leg of the triangle, then form the other leg by continuing straight along the x axis back to the origin. How long are these two legs? How long therefore is the hypotenuse?
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RESPONSE --> One side is 13, the other is 7. I found these by comparing the appropriate x or y value to its adjacent one, finding the difference. c^2=a^2+b^2 =7^2+13^2=49+169=218=sqrt(218)=14.8 14.8 is the hypothenuse length. confidence assessment: 3
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18:01:13 The leg from (7, 13) to the x axis drops from the point (7, 13) to the point (7,0) and so has length 13. The second leg runs from (7,0) to the origin, a distance of 7 units. The legs therefore have lengths a = 13 and b = 7, so that the hypotenuse c satisfies c^2 = a^2 + b^2 and we have c = `sqrt(a^2 + b^2) = `sqrt( 13^2 + 7^2 ) = `sqrt( 216) = 14.7, approximately.
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RESPONSE --> I understand. self critique assessment: 3
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18:01:55 `q002. Sketch a right triangle on a set of coordinate axes whose leg along the x axis has length 12 and whose hypotenuse has length 15. What must be the length of the second leg of the triangle?
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RESPONSE --> 15^2=12^2+b^2 225=144+b^2 81=b^2 9=b confidence assessment: 3
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18:02:01 Let c stand for the length of the hypotenuse and a for the length of the known side, with b standing for the length of the unknown side. Then since c^2 = a^2 + b^2, b^2 = c^2 - a^2 and b = `sqrt(c^2 - a^2) = `sqrt( 15^2 - 12^2) = `sqrt(225 - 144) = `sqrt(81) = 9.
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RESPONSE --> I understand. self critique assessment: 3
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18:03:08 `q003. If a line of length L is directed from the origin of an x-y coordinate system at an angle `theta with the positive x axis, then the x and y coordinates of the point where the line ends will be y = L * sin(`theta) and x = L * cos(`theta). Sketch a line of length 10, directed from the origin at an angle of 37 degrees with the positive x axis. Without doing any calculations estimate the x and y coordinates of this point. Give your results and explain how you obtained your estimates.
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RESPONSE --> (x,y)=(L cos(T), L sin (T)) I am guessing that the point is somewhere around (7, 5) simply because I know the y will be smaller than the x (since the side is long but not ""high"") and because neither are very large numbers. confidence assessment: 2
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18:03:18 The line will run closer to the x axis then to the y axis, since the line is directed at an angle below 45 degrees. It won't run a whole lot closer but it will run significantly closer, which will make the x coordinate greater than the y coordinate. Since the line itself has length 10, it will run less than 10 units along either the x or the y axis. It turns out that the x coordinate is very close to 8 and the y coordinate is very close to 6. Your estimates should have been reasonably close to these values.
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RESPONSE --> I was relatively close. self critique assessment: 2
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18:04:34 `q004. Using your calculator you can calculate sin(37 deg) and cos(37 deg). First be sure your calculator is in degree mode (this is the default mode for most calculators so if you don't know what mode your calculator is in, it is probably in degree mode). Then using the sin and cos buttons on your calculator you can find sin(37 deg) and cos(37 deg). What are these values and what should therefore be the x and y coordinates of the line directed from the origin at 37 degrees from the x axis?
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RESPONSE --> Sin(37)=.64 Cos(37)=.76 X=L Cos(T)=10*Cos(37)=7.6 Y=L Sin(T)=10*Sin(37)=6.4 confidence assessment: 3
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18:04:39 sin(37 deg) should give you a result very close but not exactly equal to .6. cos(37 deg) should give you a result very close but not exactly equal to .8. Since the x coordinate is L cos(`theta), then for L = 10 and `theta = 37 deg we have x coordinate 10 * cos(37 deg) = 10 * .8 = 8 (your result should be slightly different than this approximate value). Since the y coordinate is L sin(`theta), then for L = 10 and `theta = 37 deg we have y coordinate 10 * sin(37 deg) = 10 * .6 = 6 (your result should be slightly different than this approximate value).
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RESPONSE --> I understand. self critique assessment: 3
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18:05:43 `q005. Show that the x and y coordinates you obtained in the preceding problem in fact give the legs of a triangle whose hypotenuse is 10.
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RESPONSE --> 6.4^2+7.6^2=c^2 9.94 is close to 10 (the decimal approximations account for the discrepancy), indicating that all previously found sides are correct. confidence assessment: 3
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18:06:01 If c stands for the hypotenuse of the triangle, then if a = 8 and b = 6 are its legs we have c = `sqrt(a^2 + b^2) = `sqrt(8^2 + 6^2) = `sqrt(100) = 10. The same will hold for the more precise values you obtained in the preceding problem.
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RESPONSE --> I understand. self critique assessment: 3
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20:56:35 `q006. A vector drawn on a coordinate system is generally depicted as a line segment of a specified length directed from the origin at a specified angle with the positive x axis. The vector is traditionally drawn with an arrow on the end away from the origin. In the preceding series of problems the line segment has length 10 and was directed at 37 degrees from the positive x axis. That line segment could have been drawn with an arrow on its end, pointing away from the origin. The components of a vector consist of a vector called the x component drawn from the origin along the x axis from the origin to x coordinate L cos(`theta), and a vector called the y component drawn from the origin along the y axis to y coordinate L sin(`theta). What are the x and y components of a vector directed at an angle of 120 degrees, as measured counterclockwise from the positive x axis, and having length 30 units?
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RESPONSE --> X=Lcos(T)=30cos(120)=-15 Y=Lcos(T)=30cos(120)=26 confidence assessment: 3
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20:56:43 The x component of this vector is vector along the x axis, from the origin to x = 30 cos(120 degrees) = -15. The y component is a vector along the y axis, from the origin to y = 30 sin(120 degrees) = 26, approx.. Note that the x component is to the left and the y component upward. This is appropriate since the 120 degree angle, has measured counterclockwise from the positive x axis, takes the vector into the second quadrant, which directs it to the left and upward.
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RESPONSE --> I understand. self critique assessment: 3
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21:09:11 `q007. The angle of a vector as measured counterclockwise from the positive x axis is easily determined if the components of the vector are known. The angle is simply arctan( y component / x component ) provided the x component is positive. If the x component is negative the angle is arctan( y component / x component ) + 180 deg. If the x component is positive and the y component negative, arctan( y component / x component ) will be a negative angle, and in this case we generally add 360 degrees in order to obtain an angle between 0 and 360 degrees. The arctan, or inverse tangent tan^-1, is usually on a calculator button marked tan^-1. Find the angle and length of each of the following vectors as measured counterclockwise from the positive x axis: A vector with x component 8.7 and y component 5. A vector with x component -2.5 and y component 4.3. A vector with x component 10 and y component -17.3.
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RESPONSE --> arctan(5/8.7)=30 arctan(4.3/-2.5) + 180= 120 arctan(-17.3/10) + 360=300 5^2+8.7^2=c^2 25+76=c^2 101=c^2 10=c 4.3^2+(-2.5)^2=c^2 18+6=c^2 24=c^2 5=c (-17.3)^2+10^2=c^2 300+100=c^2 400=c^2 20=c confidence assessment: 3
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21:09:20 A vector with x component 8.7 and y component 5 makes an angle of arctan (5/8.7) = 30 degrees with the x axis. Since the x component is positive, this angle need not be modified. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(8.7^2 + 5^2) = 10. A vector with x component -2.5 and y component 4.3 makes an angle of arctan (4.3 / -2.5) + 180 deg = -60 deg + 180 deg = 120 degrees with the x axis. Since the x component is negative, we had to add the 180 degrees. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt((-2.5)^2 + 4.3^2) = 5. A vector with x component 10 and y component -17.3 makes an angle of arctan (-17.3 / 10) = -60 degrees with the x axis. Since the angle is negative, we add 360 deg to get 300 deg. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(10^2 + (-17.3)^2) = 20.
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RESPONSE --> I understand this stuff well. self critique assessment: 3
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?T???y?L?q???????assignment #019 019. Vector quantities Physics II 06-11-2007
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23:10:29 `q001. Note that this assignment contains 5 questions. . If you move 3 miles directly east then 4 miles directly north, how far do end up from your starting point and what angle would a straight line from your starting point to your ending point make relative to the eastward direction?
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RESPONSE --> 3^2+4^2=c^2 9+16=c^2 25=C^2 5=c 5 is the length arctan (4/3)=53 degrees This is the angle measurement. confidence assessment: 3
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23:10:34 If we identify the easterly direction with the positive x axis then these movements correspond to the displacement vector with x component 3 miles and y component 4 miles. This vector will have length, determined by the Pythagorean Theorem, equal to `sqrt( (3 mi)^2 + (4 mi)^2 ) = 5 miles. The angle made by this vector with the positive x axis is arctan (4 miles/(three miles)) = 53 degrees.
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RESPONSE --> I understand. self critique assessment: 3
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09:25:53 `q002. When analyzing the force is acting on an automobile as it coasts down the five degree incline, we use and x-y coordinate system with the x axis directed up the incline and the y axis perpendicular to the incline. On this 'tilted' set of axes, the 15,000 Newton weight of the car is represented by a vector acting straight downward. This puts the weight vector at an angle of 265 degrees as measured counterclockwise from the positive x axis. What are the x and y components of this weight vector? Question by student and instructor response:I have my tilted set of axes, but I cannot figure out how to graph the 15,000 N weight straight downward. Is it straight down the negative y-axis or straight down the incline? ** Neither. It is vertically downward. Since the x axis 'tilts' 5 degrees the angle between the x axis and the y axis is only 85 degrees, not 90 degrees. If we start with the x and y axes in the usual horizontal and vertical positions and rotate the axes in the counterclockwise direction until the x axis is 'tilted' 5 degrees above horizontal, then the angle between the positive x axis and the downward vertical direction will decrease from 270 deg to 265 deg. ** It might help also to magine trying to hold back a car on a hill. The force tending to accelerated the car down the hill is the component of the gravitational force which is parallel to the hill. This is the force you're trying to hold back. If the hill isn't too steep you can manage it. You couldn't hold back the entire weight of the car but you can hold back this component.
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RESPONSE --> I am still having some problems ""visualizing"" this, but I think I can figure out the x and y components fomr the information given. x=15000Ncos(265)=-1300N y=15,000Nsin(265)=-14,900N confidence assessment: 2
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09:26:34 The x component of the weight vector is 15,000 Newtons * cos (265 degrees) = -1300 Newtons approximately. The y component of the weight vector is 15,000 Newtons * sin(265 degrees) = -14,900 Newtons. Note for future reference that it is the -1300 Newtons acting in the x direction, which is parallel to the incline, then tends to make the vehicle accelerate down the incline. The -14,900 Newtons in the y direction, which is perpendicular to the incline, tend to bend or compress the incline, which if the incline is sufficiently strong causes the incline to exert a force back in the opposite direction. This force supports the automobile on the incline.
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RESPONSE --> I understand the first part and I think I get the gist of the second. self critique assessment: 3
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18:36:01 `q003. If in an attempt to move a heavy object resting on the origin of an x-y coordinate system I exert a force of 300 Newtons in the direction of the positive x axis while you exert a force of 400 Newtons in the direction of the negative y axis, then how much total force do we exert on the object and what is the direction of this force?
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RESPONSE --> a^2+b^2=c^2 (300)^2+(-400)^2=c^2 90000+160,000=c^2 500=c arctan (-400/300)=-53 degrees +360 (since y is negative)=307 degrees confidence assessment: 3
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18:36:07 Force is a vector quantity, so we can determine the magnitude of the total force using the Pythagorean Theorem. The magnitude is `sqrt( (300 N)^2 + (-400 N)^2 ) = 500 N. The angle of this force has measured counterclockwise from the positive x axis is arctan ( -400 N / (300 N) ) = -53 deg, which we express as -53 degrees + 360 degrees = 307 degrees.
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RESPONSE --> I understand. self critique assessment: 3
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19:44:54 `q004. If I exert a force of 200 Newtons an angle of 30 degrees, and you exert a force of 300 Newtons at an angle of 150 degrees, then how great will be our total force and what will be its direction?
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RESPONSE --> Mine: x=200Ncos(30)=about 173N y=200sin(30)=100N Yours: x=300Ncos(150)=-260N y=300Nsin(150)=150N x+x=173+-260=-87N y+y=100+150=250N (-87)^2+(250)^2=c^2 7569+62,500=c^2 70,069=c^2 about 265=c 265 is the magnitude of the force. The angle of the force is arctan (250N/-87N) =-71 degrees -71+180 (because x is negative)=109 degrees confidence assessment: 3
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19:45:02 My force has an x component of 200 Newtons * cosine (30 degrees) = 173 Newtons approximately, and a y component of 200 Newtons * sine (30 degrees) = 100 Newtons. This means that the action of my force is completely equivalent to the action of two forces, one of 173 Newtons in the x direction and one of 100 Newtons in the y direction. Your force has an x component of 300 Newtons * cosine (150 degrees) = -260 Newtons and a y component of 300 Newtons * sine (150 degrees) = 150 Newtons. This means that the action of your force is completely equivalent the action of two forces, one of -260 Newtons in the x direction and one of 150 Newtons in the y direction. In the x direction and we therefore have forces of 173 Newtons and -260 Newtons, which add up to a total x force of -87 Newtons. In the y direction we have forces of 100 Newtons and 150 Newtons, which add up to a total y force of 250 Newtons. The total force therefore has x component -87 Newtons and y component 250 Newtons. We easily find the magnitude and direction of this force using the Pythagorean Theorem and the arctan. The magnitude of the force is `sqrt( (-87 Newtons) ^ 2 + (250 Newtons) ^ 2) = 260 Newtons, approximately. The angle at which the force is directed, as measured counterclockwise from the positive x axis, is arctan (250 Newtons/(-87 Newtons) ) + 180 deg = -71 deg + 180 deg = 109 deg.
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RESPONSE --> I understand. self critique assessment: 3
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19:47:01 `q005. Two objects, the first with a momentum of 120 kg meters/second directed at angle 60 degrees and the second with a momentum of 80 kg meters/second directed at an angle of 330 degrees, both angles measured counterclockwise from the positive x axis, collide. What is the total momentum of the two objects before the collision?
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RESPONSE --> I am not sure how to work this. I understand sort of what I am supposed to do (the x, y thing), but I am not sure where the kg m/s go... confidence assessment: 0
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19:50:26 The momentum of the first object has x component 120 kg meters/second * cosine (60 degrees) = 60 kg meters/second and y component 120 kg meters/second * sine (60 degrees) = 103 kg meters/second. The momentum of the second object has x component 80 kg meters/second * cosine (330 degrees) = 70 kg meters/second and y component 80 kg meters/second * sine (330 degrees) = -40 kg meters/second. The total momentum therefore has x component 60 kg meters/second + 70 kg meters/second = 130 kg meters/second, and y component 103 kg meters/second + (-40 kg meters/second) = 63 kg meters/second. The magnitude of the total momentum is therefore `sqrt((130 kg meters/second) ^ 2 + (63 kg meters/second) ^ 2) = 145 kg meters/second, approximately. The direction of the total momentum makes angle arctan (63 kg meters/second / (130 kg meters/second)) = 27 degrees, approximately.
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RESPONSE --> I will be lazy and explain this in general terms instead of retyping it all. The 120kg m/s goes in front of the cos/sin, and the degrees, of course, follow the trigonometric functions. We find x and y for both objects. Then, we add both x's and both y's, getting the total x and total y for both objects. Then, we use the pythagorean theorem to find the magnitude from the sums of the x's and y's above. self critique assessment: 2
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