course Phy 201 ???&?????\z???Student Name: assignment #019
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18:25:31 `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 --> I must find the resultant displacement using the pythagorean theorem. a^2 + b^2 = c^2 9+16=c^2 c=5miles To find the angle made by the vector, I take the arctan of the y component divided by the x component. arctan (4/3) = 53degrees Since the x component is not <0, I do not have to add 180 degrees.
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18:25:42 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
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18:29:14 `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 --> We find the x component by multiplying the weight of the car by the cos of the angle. x component = 15,000N * cos(265) = -1300N We find the y component by multiplying the weight of the car by the sin of the angle. y component = 15,000N * sin(265) = -14,900n
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18:29:30 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 --> OK
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18:33:29 `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 --> I will use the pythagorean theorem to calculate the total force. a^2+b^2=c^2 300^2+-400^2 90,000+16,000=c^2 c=500N To find the angle of force, I find the arctan of the y component divided by the x component. arctan(-400/300)=-53 Since the x component is <0, I must add 180 -53+180=127degrees
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18:35:39 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 --> oops, I got the x and y confused there. So, if the y component is <0, do you add 360 degrees.
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18:35:56 `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 -->
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18:45:15 `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 --> The x component of my force would be 200N *cos(30degrees) = 173N The y component of my force would be 200N * sin(30degrees) = 100N The x component of your force would be 300N * cos(150) = -260N The y component of your force would be 300N * sin(150) = 150N In order to find the total force I must first find the sum of the forces in the x direction and that of the forces in the y direction. x direction = 173-260=-87N y direction = 100+150=250N Now, I must find the magnitude of these forces using the pythagorean theorem. 7569+62500=c^2 c=265N The angle of its direction will be arctan (250/-87) = 109degrees
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18:45:38 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 --> ok
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18:53:56 `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 --> First I will find the x and y components of the first object. x component = 120kg*m/sec * cos(60) = 60kg*m/sec y component = 120kg*m/sec * sin (60)= 103kg*m/sec Now, I wil find the x and y components of the second object. x component = 80*cos(330)=70kg*m/sec y component=80*sin(330)=-40kg*m/sec The total of the x components is 60+70=130kg*m/sec. The total of the y components is 103-40 = 63kg*m/sec Now, I must find the magnitude using the pythagorean theorem. a^2 + b^2 = c^2 16900 + 3969 = c^2 c = 144kg*m/sec
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18:54:20 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 --> oh, I forgot to add the angle of direction.
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??|????y?}???assignment #019 ???????????Physics I 07-07-2006
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18:59:12 Query class notes #20 Explain how we calculate the components of a vector given its magnitude and its angle with the positive x axis.
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RESPONSE --> The x component of a vector is calculated by multiplying the magnitude by the cos of the angle. The y component is calculated by multiplying the magnitude by the sin of the angle.
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18:59:23 ** STUDENT RESPONSE: x component of the vector = magnitude * cos of the angle y component of the vector = magnitude * sin of the angle To get the magnitude and angle from components: angle = arctan( y component / x component ); if the x component is less than 0 than we add 180 deg to the solution To get the magnitude we take the `sqrt of ( x component^2 + y component^2) **
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RESPONSE --> ok
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19:00:43 Explain what we mean when we say that the effect of a force is completely equivalent to the effect of two forces equal to its components.
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RESPONSE --> The force vector corresponding to two forces in the x and y directions is equal to the actual force exerted.
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19:01:13 ** If one person pulls with the given force F in the given direction the effect is identical to what would happen if two people pulled, one in the x direction with force Fx and the other in the y direction with force Fy. **
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RESPONSE --> okay, i dont know if my answer made much sense.
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19:02:36 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> I'm not quite sure
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19:05:48 ** Using initial conditions and the equations of motion we can determine the x and y velocities vx and vy at a given point, using the usual procedures for projectiles. The magnitude of the velocity is sqrt(vx^2 + vy^2) and the angle with the pos x axis is arctan(vy / vx), plus 180 deg if x is negative. **
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RESPONSE --> what are these usual procedures for projectiles. Would it be the equations stating Vo = magnitude * cos(theta), and Vo = magnitude * sin(theta).
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19:08:22 Explain how we can calculate the initial velocities of a projectile in the horizontal and vertical directions given the magnitude and direction of the initial velocity.
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RESPONSE --> I would assume that I would use the equations for projectiles. Initial velocity in horizontal direction = magnitude * cos(theta). Initial velocity in vertical direction = magnitude * sin(theta).
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19:08:41 ** Initial vel in the x direction is v cos(theta), where v and theta are the magnitude and the angle with respect to the positive x axis. Initial vel in the y direction is v sin(theta). **
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RESPONSE --> ok
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