course Phy 201 Here is my work for Asst 19. I know it may be a little shotty but I need to do some more work on the class notes to understand this in greater detail. I did get your messages on all the work and thank you for getting back to me. The reason I did that is because it had been about three days since I heard anything from you. Sorry about dumping all of that on you.
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20:37:43 `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 --> This forms a right triangle when you draw the straight line from your starting point to where you end up. So you can say: c^2 = a^2 + b^2, or c =sqrt ( ( 3^2 ) + ( 4^2 ) c = 5 And the straight line is the hypotenuse of the triangle.
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20:41:26 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 I didn't see where you asked for the angle of the triangle, but maybe that is where you asked how is the straight line relative to the eastward direction. Any how I know how to get this. You simply set the calculator to tan-1 and then divide 4 by 3 to get the 53 degrees.
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20:52:16 `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 very lost as to how to set this up?
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20:55:42 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 see what you did but I still don't really understand at the moment. I know that what you did multiply the L or leg by the sin or cos of the angle. What I don't know is how it is 265 degrees and not 5 degrees? And how the Force of the weight is the L?
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21:02:24 `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 --> Well the way I see it that we exert a total force of 700 Newtons on the object. Even if there is 300 going in the positive direction and 400 going in the negative direction there is still a total of 700 Newtons exerted on it. Now since there is 100 Newtons more of negative force exerted than positive then I would say that the object is now moving in a negative direction or in the direction of the y axis.
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21:07:26 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 --> Ok so if we say that force is a vector quantity then that is why we can treat it as if it were the length of a triangle leg. This answers my question from the last self critique. I guess this throws me off because I am looking for just a straight forward response to something like "" It was moving in the negative direction"" and not 307 degrees.
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21:14:40 `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 --> Ok I think that you could use the Pythagorean Theorem and say : c = sqrt ( ( 200^2) + ( 300^2)
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21:16:56 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 --> Sorry but this will take me some time to absorb. I understand all of the calculations but how to set it up at first is really throwing me for a loop right now.
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21:26:28 `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 don't know why but I will set up the two using the sin cos method. 120 kg m/s * cos(60 deg) = x = 60 120 kg m/s * sin (60 deg) = y = 104 80 kg m/s * cos ( 330 deg) = x = 69.3 80 kg m/s * sin ( 330 deg) = y = -40 So the total of the x axis is 164 and the y axis is 29.3.
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21:29:32 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 --> First of all I didn't add the right components and secondly I didn't go the whole way to get the angle of the two. Needless to say this is going to take a little work and I am not sharp on it at this point.
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ñÀOñ°~ÂÃó ‘‘MÏèE¸¬‹šãæaó™¸Í assignment #019 yEß܃øŽ¶Œª³á¿ãŒêÝŒüÁÖ Physics I 12-05-2005
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21:41:02 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 --> We would simply take the magnitude and the vector and set up: L * cos(angle) = x axis L * sin(angle) = y axis
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21:42:52 ** 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 --> I forgot once again to add in the way to get the angle. That is if the x axis is negative then add 180 degrees, if x is positive then add nothing and if y is negative then add 360 degrees.
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21:44:45 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 --> Well in my terms it means that the effect of a force like the angle and magnitude that was in the last question is equal to the components that come out of its solution.
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21:44:56 ** 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 -->
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21:52:02 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> If we know one given point then the other would be going at a rate of 9.8 m/s in the negative or positive direction. Then you can work out the other leg by way of the Pythagorean Theorem.
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21:52:31 ** 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 -->
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21:54:24 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 like to say the same way that we did the last problem but that is probably wrong.
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21:55:50 ** 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 these are all things that I have done before I just don't know how to figure the answers by the wording that is being used. I guess I am not ready for the terminology yet.
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21:55:54 Univ. 8.58 (8.56 10th edition). 40 g, dropped from 2.00 m, rebounds to 1.60 m, .200 ms contact. Impulse? Ave. force?
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RESPONSE -->
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21:55:56 ** You have to find the momentum of the ball immediately before and immediately after the encounter with the floor. This allows you to find change in momentum. Using downward as positive direction throughout: Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.). It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This gives rebound velocity v0 = -5.6 m/s approx. Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is about .04 kg * -11.9 m/s = -.48 kg m/s. In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s / (.002 s) = -240 Newtons, approx. **
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RESPONSE -->
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21:56:57 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I really need to look at the notes in greater detail after this in order to understand the lesson to a point of acceptable proficiency.
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