course Phy 121 013. `query 13 Question: `qprin 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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Given Solution: `aA 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. Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qprin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel Your solution: Vavg=125/2=62.5m/s .7m/62.5=.0112s. A=125/.0112=11160m/s^2 F=.007*11160=78N confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The initial velocity of the bullet is zero and the final velocity is 125 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) / (.011 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qSTUDENT Self-critique rating #$&* ********************************************* Question: 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. Your solution: Fnet=ma=4.5*m Fnet+m=T-mg T=m(a+g) 22=m(4.5+9.8)=1.5kg Confidence rating #$&*: 3
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Given Solution: `a** Think in terms of net force. The net force on the fish must be Fnet = m a = m * 4.5 m/s^2. Net force is tension + weight = T - m g, assuming the upward direction is positive. So T - m g = m a and T = m a + m g. Factoring out m we have T = m ( a + g ) so that m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK "