course phy201 |????????????assignment #019019. `query 19
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22:26:00 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 --> Xcompt = magnitude(cos theta) Y compt. = magnitude(sin theta) R = sqrt x^2 + y^2 confidence assessment: 3
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22:26:37 ** 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:
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22:28:01 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 --> these two forces are equal to the single force but shows the magnitude and direction of the object. The components show the alternate forces = to the original confidence assessment: 2
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22:28:13 ** 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 --> self critique assessment:
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22:28:16 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> confidence assessment:
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22:30:07 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> To find the velocity we must have dt which we find from using the v0 of the vertical motion since we have more known variables in this direction. After finding dt we can then find the components of the horizontal motion hence velocity confidence assessment: 2
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22:32:07 ** 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 --> R = sqrt (total x compon)^2 + (total y compn)^2 arctan ycomp / x comp = theta add 360 if y is - add 180 is x is - self critique assessment: 2
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22:34:08 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 --> break the motion into seperate x and y components taking into account an angle if there is one present confidence assessment: 2
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22:34:33 ** 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 --> self critique assessment:
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22:41:03 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 --> Fnet * ds = mvf - mv0 Fnet * ds = impulse Fnet = 40 * 9.8 = 392 * 2m impulse1 = 784 Fnet = -392 * 1.6 = -627 Total imp = 784 - 627 = 157 force = 392 confidence assessment: 1
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23:01:05 ** 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 --> The first thing I did wrong was forget to convert. 40g = .04kg .200ms = .002s In order to find Impulse I needed to find momentum which required me to determine Vo and Vf for the two motions; falling and then rebound. v0 = 0 a = 9.8 ds = 2m vf^2 = 0 + 2 * 9.8 * 2 vf = 6.3m/s rebound vf = 0 a = 9.8 ds =-1.6 v0^2 = 2 *9.8 * -1.6 - 0 v0 = -5.6 I understand that 1.6 is - because it is going in the opps. direction of the original motion. I dont understand why you would subtract the two and the drop velocity; vf 6.3 would be placed second in the equation. Fnet * dt = mvf - mv0 p = -5.6 -6.3 = -11.9 * .04kg p = -.48kg m/s/s = N Fnet * dt = dp Fnet = .48kgm/s / .002s Fnet = 240 kg m/s/ self critique assessment: 3
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23:01:11 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment:
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?????????????assignment #020 020. `query 20 Physics I 04-17-2009
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23:49:16 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
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RESPONSE --> we divide the single force into two forces that would equal the original confidence assessment: 3
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23:50:12 ** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. **
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RESPONSE --> R = sqrt totalxcom^2 + totalycomp^2 from the pyth therom a^2 + b^2 = c^2 self critique assessment: 2
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23:51:49 Explain how we get the components of a vector from its angle and magnitude.
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RESPONSE --> seperate the x and y components for the vector based on its angle. x cos theta y sin theta confidence assessment: 3
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23:52:09 ** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **
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RESPONSE --> self critique assessment:
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23:55:17 prin phy, gen phy 7.02. Const frict force of 25 N on a 65 kg skiier for 20 sec; what is change in vel?
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RESPONSE --> F = 25N m = 65kg dt = 20sec dv = ? Fnet * dt = dp Fnet * ds = mv 25*20/60 = 8.3m/s confidence assessment: 2
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23:58:28 If the direction of the velocity is taken to be positive, then the directio of the frictional force is negative. A constant frictional force of -25 N for 20 sec delivers an impulse of -25 N * 20 sec = -500 N sec in the direction opposite the velocity. By the impulse-momentum theorem we have impulse = change in momentum, so the change in momentum must be -500 N sec. The change in momentum is m * `dv, so we have m `dv = impulse and `dv = impulse / m = -500 N s / (65 kg) = -7.7 N s / kg = -7.7 kg m/s^2 * s / kg = -7.7 m/s.
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RESPONSE --> I used 60kg as the mass instead of 65kg however I understood which equations to use and why but I ignored the fact friction is a resisting force and would obviously be -. Normally when i double check my work on test and i tend to catch and correct these simple mistakes self critique assessment: 3
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00:00:24 gen phy #7.12 23 g bullet 230 m/s 2-kg block emerges at 170 m/s speed of block
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RESPONSE --> I dont understand this question. m = 23 v =230m/s / 2kg block confidence assessment: 0
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00:01:16 **STUDENT SOLUTION: Momentum conservation gives us m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have (.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v). Solving for v: (5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v) .78kg m/s = 2kg * v v = 1.38 kg m/s / (2 kg) = .69 m/s. INSTRUCTOR COMMENT: It's probably easier to solve for the variable v2 ': Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2. Substituting for m1, v1, m2, v2 we will get the result you obtained.**
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RESPONSE --> self critique assessment:
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00:02:16 **** Univ. 8.70 (8.68 10th edition). 8 g bullet into .992 kg block, compresses spring 15 cm. .75 N force compresses .25 cm. Vel of block just after impact, vel of bullet?
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RESPONSE --> confidence assessment:
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00:02:59 ** The spring ideally obeys Hook's Law F = k x. It follows that k = .75 N / .25 cm = 3 N / cm. At compression 15 cm the potential energy of the system is PE = .5 k x^2 = .5 * 3 N/cm * (15 cm)^2 = 337.5 N cm, or 3.375 N m = 3.375 Joules, which we round to three significant figures to get 3.38 J. The KE of the 1 kg mass (block + bullet) just after impact is in the ideal case all converted to this PE so the velocity of the block was v such that .5 m v^2 = PE, or v = sqrt(2 PE / m) = sqrt(2 * 3.38 J / ( 1 kg) ) = 2.6 m/s, approx. The momentum of the 1 kg mass was therefore 2.6 m/s * .992 kg = 2.6 kg m/s, approx., just after collision with the bullet. Just before collision the momentum of the block was zero so by conservation of momentum the momentum of the bullet was 2.6 kg m/s. So we have mBullet * vBullet = 2.6 kg m/s and vBullet = 2.6 kg m/s / (.008 kg) = 330 m/s, approx. **
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RESPONSE --> self critique assessment:
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