course Phy 121 assignment #013013. `query 13
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22:07:12 prin phy and gen phy problem 4.02 net force 265 N on bike and rider accelerates at 2.30 m/s^2, mass of bike and rider
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RESPONSE --> m = 265N/2.30m/s^2 = 115kg confidence assessment: 2
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22:07:30 A force Fnet acting on mass m results in acceleration a, where a = Fnet / m. We are given Fnet and a, so we can solve the equation to find m. Multiplying both sides by m we get a * m = Fnet / m * m so a * m = Fnet. Dividing both sides of this equation by a we have m = Fnet / a = 265 N / (2.30 m/s^2) = 115 (kg m/s^2) / (m/s^2) = 115 kg.
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RESPONSE --> self critique assessment: 3
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22:17:52 prin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel
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RESPONSE --> 7g pellet * 1g/1000kg = .007kg force needed = .007kg * 125m/s = .875 Newtons
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22:18:15 ** The initial velocity of the bullet is zero and the final velocity is 175 m/s. If we assume uniform acceleration (not necessarily the case but not a bad first approximation) the average velocity is (0 + 125 m/s) / 2 = 62.5 m/s and the time required for the trip down the barrel is .7 m / (62.5 m/s) = .011 sec, approx.. Acceleration is therefore rate of velocity change = `dv / `dt = (125 m/s - 0 m/s) / (.11 sec) = 11000 m/s^2, approx.. The force on the bullet is therefore F = m a = .007 kg * 11000 m/s^2 = 77 N approx. **
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RESPONSE --> Understand self critique assessment: 3
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22:19:05 ** Weight is force exerted by gravity. Net force is Fnet = m * a. The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity. So m a = 50 N - m g, which we solve for m to get m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg. If the balance reads 30 N then Fnet is still m * a and we have m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2. If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2. Net force will be -m g; net force is also Fbalance - m g. So -m g = Fbalance - m g and we conclude that the balance exerts no force. So it reads 0. **
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RESPONSE --> self critique assessment: 3
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22:19:35 ** Think in terms of net force. The net force on the fish must be Fnet = m a = m * 4.5 m/s^2. Net force is tension + weight = T - m g, assuming the upward direction is positive. So T - m g = m a and T = m a + m g. Factoring out m we have T = m ( a + g ) so that m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx.. The same principles apply with the elevator. **
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RESPONSE --> OK self critique assessment: 1
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