7.12 Unit Test Part 2

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Ohio University, Main Campus *

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2001

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Physics

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Dec 6, 2023

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3

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Name: Click or tap here to enter text. Date: Click or tap here to enter text. Graded Assignment Unit Test, Part 2 Answer the questions, using complete sentences. When you have finished, submit this assignment to your teacher by the due date for full credit. Total score: of 25 points (Score for Question 1: of 8 points) During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of at an angle of 50.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work. Answer: The projectile may fly up to 26.4 meters in the air. 1) Time of takeoff: T= 2usin θ /g u= initial velocity θ = angle of projection g= acceleration due to gravity T= 2 * 30.0 * sin 50.0°/10 T= 4.6s 2) The range; R=u^2sin2 θ /g R=(30.0)^2 * sin 2( sin 50.0°)/10 R=88.6 m 3) The max height; H=u^2sin^2 θ /2g H=(30.0)^2 * sin^2(50.0°)/2 * 10 H=26.4 m Click or tap here to enter text. (Score for Question 2: of 8 points) A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 10.0 m) and spun at a constant angular velocity of 16.3 rpm. Answer the following: What is the angular velocity of the centrifuge in ? What is the linear velocity of the astronaut at the outer edge of the centrifuge? What is the centripetal acceleration of the astronaut at the end of the centrifuge? How many g ’s does the astronaut experience? What is the centripetal force and net torque experienced by the astronaut? Give magnitudes and directions. Show your work. Answer: 1.3823 rad/s 20.7345 m/s
28.66129935 m/s^2 a= 2.92164g 2006.29095 N In a radial motion My work: r = Radius = 15 m m = Mass of person = 70 kg g = Gravitational acceleration = 9.81 m/s^2 Angle velocity is determined by w=13.2 * 2 π /60 -> w=1.3823 rad/s Angular velocity is 1.3823 rad/s The formula for linear velocity is v=rw -> v =15 * 1.3823 -> v = 20.7345 m/s The linear velocity is 20.7345 m/s The source of centripetal acceleration is ac = rw^2 -> ac = 15 * 1.3823^2 -> ac = 28.66129935 m/s^ . The centripetal acceleration is 28.66129935 m/s ² acceleration expressed in g a/g = 28.66129935/9.81 -> a = 2.92164g a = 2.92164g The source of centripetal force is Fc = mac -> Fc = 70 * 28.66129935 -> Fc = 2006.29095 N (Score for Question 3: of 9 points) An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion, as shown in this graph. Answer the following questions: What is the period and frequency of the pendulum’s motion? How many seconds out of phase with the displacements shown would graphs of the velocity and acceleration be? What is the acceleration due to gravity on the surface of the planet in ? Determine the number of g- forces. Show any necessary calculations. Answer: 1)I can count 9 full cycles in 5 seconds from the graph. The frequency is therefore 9/5, or 1.8 Hz, or once every second. 1/frequency = 5 seconds / 9 cycles = 0.555 seconds is the period. 2)Velocity is the first derivative of displacement and is 1/4 cycle behind, or 5/36 seconds behind displacement, or 0.13888 seconds. Acceleration is the first derivative of velocity and the second derivative of displacement. It is 1/4 cycle behind velocity and 1/2 cycle behind displacement. It is 5/36 (0.13888...) seconds behind velocity and 5/18
(0.2777...) seconds behind displacement. 3)The period of a perfect pendulum anywhere is for small swing angles T=2pi sqrt(length/local gravity) We can use the astronaut's data and this formula to determine the local gravity where he is because he has already performed the pendulum and relayed the data to us. P = 2pi sqrt(length/local gravity) 5/9 sec = 2 π sqrt(0.2m / gravity) sqrt(0.2m / gravity) = 5/9 sec / 2 π Take each side's reciprocal: (gravity) / (0.2) = 18 π / 5 Multiply each side by (0.2): (gravity) = 18 π (0.2) / 5 Square both sides: Gravity = (324 π² ) (0.2) / 25 = 25.582 m/sec ² This is equivalent to 2.608 times the gravity of the Earth. Therefore, it should not come as a surprise that the astronaut returned to his landing capsule to bed down after only 5 seconds of toying with his pendulum
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