course PHY 201 ??~????????b???Sx?assignment #019
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12:42:33 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 --> A vector can be broken into its component x and y pieces by using the angle and magnitude and finding the ration between x and y via the following equation: x = L * cos(theta) y = L * sin (theta) confidence assessment: 2
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12:42:46 ** 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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12:44:56 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 --> Each force can be decomposed into an x and y component, and by summing these components we acheive again the original force. The result of the original force and the result of the summed component forces are exactly identical. confidence assessment: 2
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12:45:03 ** 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 --> self critique assessment: 3
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12:57:17 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> By using the equations for determining single magnitude and direction from an x velocity and a y velocity: magnitude = sqrt( x^2 + y^2) angle = arctan(y/x) + 180 if x is negative or + 360 if y is negative confidence assessment: 2
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13:03:52 ** 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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13:05:53 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 --> By breaking the given momentum into component velocities for the x and y axis using the following equations: x = L * cos(theta) y = L * sin(theta) confidence assessment: 2
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13:06:14 ** 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 --> self critique assessment: 3
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13:06:26 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 --> not univ phy confidence assessment: 2
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13:06:31 ** 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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13:06:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 3
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