course Phy 201 ŤšĹy¤„´Ü‘˛J¬{„Şüń‘Ůą‹’ćassignment #020
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10:58:53 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
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RESPONSE --> The resultant is the sum of the original vectors. Therefore, we add all the x components, to get the x component of the resultant, and add all the y components to get the y component of the resultant confidence assessment: 3
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10:59:08 ** 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: 3
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10:59:52 Explain how we get the components of a vector from its angle and magnitude.
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RESPONSE --> x component = magnitude *cos(angle) y component = magnitude *sin(angle) confidence assessment: 3
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11:00:01 ** 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: 3
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11:23:49 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 --> I was using the formula Fnet = m`dv / `dt, but I was using the weight from problem 7.01, not 7.02, and I was getting a really large change in velocity. After I saw in the answer the weight is 65kg, I immediately made the change and got the correct answer. Although your way seems shorter. self critique assessment: 2
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11:33:43 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 --> let A represent bullet, B represent block pA + pB = p`A + p`B, therefore, mAvA + mBvB = mAv`A + mBv`B .023kg(230m/s) + 2kg(0m/s) = .023kg(170m/s) + 2kgv`B 5.29kgm/s = 3.91kgm/s + 2kg*v`B 1.38kg m/s = 2kg*v`B .69m/s = v`B Velocity of the block after collision is .69 m/s confidence assessment: 3
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11:35:44 **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: 3
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11:35:58 **** 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 --> N/A confidence assessment: 0
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