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13:35:24 `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 --> We make the 3miles the x-axis and 4miles the y-axis. Since we are moving in the east direction x will be positive. First we use the pythag. theorem to find the length. c=sqrt(3^2miles+4^2miles),c=sqrt(25),c=5miles. The angle of the vector will be (y/x)tan^-1=deg.We have (4miles/3miles)tan^-1=53.1deg confidence assessment: 3
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13:35:42 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 --> ok self critique assessment: 3
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13:45:06 `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 --> If you take the vertical downward force at the 270deg mark and we start with the 5deg incline then we have difference 265 degrees. If we use the formula L cos(theta) and sin we can find the x,y component. 15000newtons*cos(265),x=-1307newtons and y=15000newtons*sin(265)=-14942newtons confidence assessment: 3
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13:47:41 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 messed the flow of this query up or got it out of whack so I did not get to answer this one self critique assessment: 2
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14:04:51 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 --> If we take I force and use the L*cos(theta) and the sin we can find the x,y components.200newtons*cos(30deg)=173=x and 200newtons*sin(30deg)=100deg. The you force is done the same way. 300newtons*cos(150)=-260degrees=x and 300newtons*sin(150)=150=y. If we add the tatal x,y forces together we can find the total force. 173new+-260new=-87newtons and y=100new=150new=250newtons.We now use the pythag. theorem to find the magnitude which is c=sqrt(-87^2+250^2),c=265newtons. Now we can find the angle by using (250newtons/-87newtons)tan^-1=-71degrees. Since we have a -x we add 180 degrees=109degrees self critique assessment: 3
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16:35:47 `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 --> We can do this the same way as the last problem. 1st object x,y components are 120kg m/s*cos(60deg)=60kg m/s and 120hg m/s*sin(60deg)=104kg m/s. 2nd object is 80kg m/s*cos(330deg)=69kg m/s and 80kg m/s*sin(330deg)=-40kg m/s. The total x-component is 60kg m/s+69kg m/s=130kg m/s and the y-component is 104kg m/s+-40kg m/s=64kg m/s.Now we know the x,y component we use the pythag. theorem c=sqrt(130kg m/s^2+64kg m/s^2),c=145kg m/s.Now we take the tan^-1 of y/x to find the angle. (64kg m/s/130kg m/s_tan^-1=26deg. The total momentum is =x component 130kg m/s and the y component is=64kg m/s. confidence assessment: 2
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16:36:50 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 --> ok self critique assessment: 3
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