course Phy 231 I am assuming that it is too late to actually finish the course. I just haven't been able to consistently motivate myself. However I am going to continue doing and submitting the work. If you have a chance to comment, great. If not, I completely understand.
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11:54:23 Query gen phy problem 4.08 force to accelerate 7 g pellet to 175 m/s in .7 m barrel
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11:54:25 ** 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 + 175 m/s) / 2 = 87.5 m/s and the time required for the trip down the barrel is .7 m / (87.5 m/s) = .008 sec, approx.. Acceleration is therefore rate of velocity change = `dv / `dt = (175 m/s - 0 m/s) / (.08 sec) = 22000 m/s^2, approx.. The force on the bullet is therefore F = m a = .007 kg * 22000 m/s^2 = 154 N approx. **
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12:01:18 univ phy problem 4.38 (4.34 10th edition) fish on balance, reading 60 N when accel is 2.45 m/s^2 up; find true weight, situation when balance reads 30 N, balance reading when cable breaks What is the net force on the fish when the balance reads 60 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read when the cable breaks?
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RESPONSE --> Actual 4.38: A parachutist relies on air resistance (mainly on her parachute) to decrease her downward velocity. She and her parachute have a mass of 55.0 kg and air resistance exerts a total upward force of 620 N on her and her parachute. a) What is the weight of the parachutist? b) Draw a free-body diagram for the parachutist. Use that diagram to calculate the net force on the parachutist. Is the net force upward or downward? c) What is the acceleration (magnitude and direction) of the parachutist? a) w = mg = 55.0 kg * 9.8 m/s^2 = 539 N b) The force of the air resistance acts upward, while the force of gravity acts downward. Fnet = 539 N + (-620 N) = -81 N (upward force of 81 N) c) Fnet = ma, so a = -81 N/55.0 kg = -1.47 m/s^2 (or 1.47 m/s^2 in an upward direction, ie slowing down)
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12:01:21 ** 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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12:01:33 STUDENT QUESTION: I had trouble with the problems involving tension in lines. For example the Fish prob. Prob#9 A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks, What is the mass of the fish. Here's what I did. Sum of F = Fup + F down -22 N = 4.5 m/s^2 * m(fish) - 9.8 m/s^2 * m(fish) -22N = -5.3 m/s^2 m(fish) m(fish) = 4.2 kg I know its wrong, I just don't know what to do.I had the same problem with the elevator tension on problem 17.
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12:01:44 ** 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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