course phy 121 öº·²í›ßë|¢ôÛœ†”ÄõáÖØ¥Ãassignment #019
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13:56:48 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 --> x = magnitude * cos (angle) y = magnitude * sin (angle) If x is negative add 180 to the answer. If y is negative add 360 to the answer. Not sure if both are negative... confidence assessment: 3
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13:57:10 ** 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 --> self critique assessment: 3
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14:01:04 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 --> I don't know how to explain it without just using an example. If you have a force with x and y components (10, 16) respectively. It is the equivalent of two seperate forces with components that add up to (10, 16). For example 1 force has x,y components (4,12) and the second force has x,y components (6,4). When you add their x's you get 10 and when you add their y's you get 16. confidence assessment: 1
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16:27:35 ** 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 --> Yeah, that's what I meant...It does makes sense. A very old example would be the Erie Canal. Two mules on opposite sides of the canal pulling a boat. They are pulling at angles to the boat and opposite each other, but the boat goes straight up the canal. And if one mule is stronger than the other, then he has to be pulling at a different angle than the weaker one to make the boat go straight. self critique assessment: 2
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18:02:57 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> The square root of x^2 +y^2 equals the magnitude. The direction is calculated by finding arctan (y/x) and THIS is where we add the 180 if x is negative and 360 is y is negative. confidence assessment: 2
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18:03:12 ** 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 --> self critique assessment: 3
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18:17:47 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 am skeptical that it is this easy... You find the x and y components of the initial velocity. x will equal the horizontal velocity and y will equal the vertical velocity. confidence assessment: 0
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18:18:33 ** 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 --> Wow! that's what I meant by finding the x and y components. self critique assessment: 3
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18:18:42 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 --> NA confidence assessment: 3
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18:18:48 ** 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 --> self critique assessment: 3
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18:24:16 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Section 7-2 of the reading surprised me. The Falling on or off a sled example. I was surprised that falling vertically onto a sled would slow it down, experience has shown me that the sled speeds up, but the reason is that I was moving in a horizontal motion first and not directly vertically. self critique assessment: 3
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