course phys201 ֣~yƷπassignment #020
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16:23:21 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
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RESPONSE --> add x component of the original vectors to get the x component of the resultant. add y component of the original vectors to get the y component of the resultant.. confidence assessment: 2
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16:23:24 ** 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 --> self critique assessment: 2
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16:23:57 Explain how we get the components of a vector from its angle and magnitude.
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RESPONSE --> y comp = sin angle of vecotr x comp = cosin angle of vector confidence assessment: 2
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16:24:36 ** 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: 2
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16:26:17 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 --> -25 N * 20 s = 500 N m * dv = impulse and dv = impulse/m = -500/65 kg = -7.7 N g/ kg = m/s^2* s/kg = m/s confidence assessment: 2
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16:30:49 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 --> self critique assessment: 2
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16:33:17 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 --> m1v1+m2v2=m1v1+m2v2 (.023 kg) .230 m/s + 2kg (0) = .023 kg (170 m/s) + 2 kg V 5.29 m/s = 3.91 kg m/s +2 kg V V= 1.38 m/s /2kg = .69 m/s confidence assessment: 2
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16:35:25 **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: 2
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16:36:30 **** 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: 2
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16:36:34 ** 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: 2
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