course Phys 201 ¾Èü›¶¹£ŸI¤Fâò닇~¶[¢ÐÑ~assignment #019
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13:28:37 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 --> When calculating the vecotr of the x axis, use M*cosin(deg) When calculatin the vector of the y axis, use M*sin(deg) to get the M and the direction of the force, use tan^-1(y axis/x axis) *add 180 if less than 0 M= sqrt. ((x axis )^2+(y axis)^2) confidence assessment: 3
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13:28:54 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 understand this question confidence assessment: 3
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13:29:03 ** 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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13:35:27 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> we use, M= sqrt.((x axis)^2+(y axis)^2) and tan^-1= (y axis/ x axis) add 180 if less 0 confidence assessment: 3
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13:45:10 ** 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 --> i understand this question self critique assessment: 3
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13:47:47 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 --> x direction = v*cosine(deg) y direction= v*sin(deg confidence assessment: 3
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13:47:58 ** 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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13:48:04 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 --> confidence assessment: 3
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13:48:09 ** 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: 0
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13:48:20 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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