Physics 20 Unit C Assessment

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Bow Valley College, Calgary *

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Jan 9, 2024

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Physics 20: Unit C Circular Motion, Work and Energy /28 At the beginning of this unit, you were given some “I can” statements and other “Questions” as part of “Questions You Should Be Able to Answer For This Unit” document. Your task for this assessment is to find an example question (from any source) that requires the information in each “Question” listed in that document. You’ll then explain how to answer each question, showing all your work, and ultimately, answer the question. You’ll be graded on each question with the following criteria: 1 point Explain how the question relates to the “Questions You Should Be Able to Answer For This Unit” question. 1 point Correct final answer for the question 2 points Relevant formula and work shown, or relevant concepts explained for how to get the answer Some questions might answer multiple “Questions” in this document. While you can use the same question more than once, you will need to re-explain how the question answers that addresses “Question” Remember, this assignment is worth 20% of your grade for this unit, so put a considerable amount of effort into this assignment.

1. Why is the acceleration of uniform circular motion directed toward the center of a circle? If a student runs at a speed of 10 m/s on a circular track of radius 240 m, what is his acceleration? R=240m V=10m/s A=? A=v^2/r A=(10m/s)^2/240m A=0.42m/s^2 The student acceleration is 0.42m/s^2. The acceleration of uniform circular motion is directed toward the center of the circle because it is necessary to continuously change the direction of the velocity vector. According to Newton's first law of motion, an object in motion will keep moving at constant speed in a straight line unless it is acted upon by an external force. We can see this in uniform circular motion, as the object moves away from the straight line and moves in a circular trajectory because of an external centripetal force acting on it. In this sample question it asks to find the centripetal acceleration of a student running on a circular track. Using the formula a = v^2/R, the acceleration can be calculated. After plugging in the values, the answer was approximately 0.42 m/s2. This sample question gives an excellent display of how the velocity and radius affect the centripetal acceleration of an object or person. Therefore, the question from the document as well as the I Can Statement are answered.

2. Using circular motion to approximate elliptical orbits, how can you describe the motion of planets, moons (natural satellites) and satellites (artificial satellites) using mathematical relationships/formula(s)? Planet Ganme has 9.00 x 104 seconds/day and 310.0 days in a year. Determine the period of the other planets (in days) from Ganme's year, using Kepler's third law. The average orbital period of Tanhe in days is… T^2=k x a^3 A=8.48x10^10m K=4.7377929 x 10^-29 T=? T=square root of k x a^3 T=square root(4.7377929 x 10^-29)(8.48x10^10m) T=169.9 T=170 days approximately The average orbital period of tanhe is 170 days. Using circular motion, we can approximate elliptical orbits and describe the motion of planets in various different ways. In elliptical orbits, planets and satellites move around another object in an oval shape or in an elliptical shape. An orbital period of a satellite or planet is the amount of time it takes it to go around the other object completely. We can find the orbital period of any object in orbit by using the formula: T^2 = k*a^3. We can also describe the motion of planets and satellites by finding their average orbital radius, which allows us to determine the shape and the path of the orbit. This question asks to find the average orbital period of one of the planets in days. To find this we can use the formula for the orbital period and we end up with an answer of approximately 170 days. This sample question allows us to connect the average orbital period with the radius and it allows us to describe how planets orbit around a main object. Therefore, the I Can Statement and the question from the document are answered.

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3. How was Newton able to use Kepler’s laws that described the motion of planets as elliptical orbits around the sun, to develop his law of universal gravitation? Using the provided scale and the length of the semi-major axis, what is the value of Tanhe's average orbital radius? Give your answer in the form of a.bc x 1010 m, you are giving the value of a.bo For example, if your answer is 1.23 x 1010 m, then you should enter 1.23 and your answer. T^2=ka^3 a = 8.48x10^10m/s K= 4.7377929 x 10^29 T=? T= square root k x a^3 T= square root ( 4.7377929 x 10^29)(8.48x10^10m/s) T=169.9= 170 days The average orbital period is 170 days approximately Kepler's law defines the motion of planets and how they orbit in a system. Because of this, Newton realized that all motion no matter what type, followed the same basic principles of gravity and motion. This helped Newton develop his law of universal gravitation because he understood that in any system, gravity is key and it is crucial for orbiting planets. This sample question allows us to see how the plant orbits in its path, and how the orbital radius plays a role in it. This question asks us to find the average orbital radius of a planet. All we have to do for this question is convert the units from cm to m and find the radius. After converting them by multiplying them by the scale, we must divide the answer by 2 to get the radius. This question allows us to see how the orbital radius of a planet relates to its motion and orbital path. Therefore, the I Can Statement and the question from the document are answered.

4. What is the relationship between the kinetic energy(due to motion), the potential energy (stored) and the total mechanical energy of an object in a closed (isolated) system? A child, initially at rest at the top of 3.55 m high slide, slides down to a height of 0.480 m. At the bottom of the slide, the child has a speed of 2.45 m/s. If the mass of the child is 34.5 kg, what is the change in mechanical energy of the child Hi=3.55m Hf=0.480m Vf=2.45m/s M=34.5kg G=9.81m/s^2 Angle E=? Ep1=mgh Ep1=(34.5kg)(9.81m/s^2)(4.03m) Ep1=1363.93335 Em=1/2mv Em=1/2(34.5)(2.45)^2 Em=103.543125 Angle E =Em-Ep E=(103.543125)-(1363.93335) E=-1260.39j The change in mechanical energy is approximately-1260.4J Kinetic energy is the energy an object has because of its motion. Anything that is moving has kinetic energy. Potential energy on the other hand, is the energy that is stored within an object because of its position relative to other objects. Mechanical energy is the energy that moves an object. These 3 are related because mechanical energy is the sum of kinetic energy and potential energy within a system. Both kinetic and potential energy are required for mechanical energy. This sample question portrays this as it asks to find the change in mechanical energy of the child. To do this we first have to find the initial potential energy and the final kinetic energy. Then subtracting the two values will give us the change in mechanical energy. This question displays how both kinetic and potential energy are needed for mechanical energy and it also shows how these 3 different types of energies are related. Therefore, the question from the document and the I Can Statement are answered.

5. What is work? A force applied to a 25 kg mass causes an acceleration of 3.5 m/s2 over 25 m. The work done on the object is M=25kg A=3.5m/s^2 D=25 F=? W=? F=ma F=(25kg)(3.5m/s)^2 F=87.5N W=fd W=(87.5N)(25) W=2187.5J=2.2 x 10^2 J Work is the measure of energy transfer that takes place when an object is moved over a distance by an external force. It is measured in joules. It is a scalar quantity and not a vector,so it only includes magnitude and not direction. Work can also be defined as the action that is due to a force causing an object to move. This sample question asks to find the work done on the object. To solve this question we can use the formula W = F*d. we must first find the force and once we have both the force and the distance we can plug our values in. After plugging in the values, the answer was 2.2*10^3 J. This question shows how work is found and how it relates to the force and distance of an object. Therefore, the question from the document and I Can Statement a

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6. How does the mechanical energy change in a system that is not isolated? A child, initially at rest at the top of 3.55 m high slide, slides down to a height of 0.480 m. At the bottom of the slide, the child has a speed of 2.45 m/s. If the mass of the child is 34.5 kg, what is the change in mechanical energy of the child Hi=3.55m Hf=0.480m Vf=2.45m/s M=34.5kg G=9.81m/s^2 Angle E=? Ep1=mgh Ep1=(34.5kg)(9.81m/s^2)(4.03m) Ep1=1363.93335 Em=1/2mv Em=1/2(34.5)(2.45)^2 Em=103.543125 Angle E =Em-Ep E=(103.543125)-(1363.93335) E=-1260.39j The change in mechanical energy is approximately-1260.4J In an isolated system, no energy can be exchanged with the surroundings or environment. However, when in a non-isolated system, energy can be exchanged with the environment and the objects surroundings. Components like friction are present in non-isolated systems and affect the energy. In this sample question, it involves a child sliding down a slide. This is an example of a non- isolated system because in reality, friction would play a large role in the energy. The change in mechanical energy would also be different if the friction of the slide was taken into account. Therefore, the question from the document is answered.

7. What is power and how does it relate to work? If a 70.0 kg student runs up a flight of stairs 15.0 m high in 25.0 s, what is the average power? M=20.0kg H=15.0m T=25.0s W=mgh W=(20.0kg)(9.81m/s^2) (15.0m) W=10300.5J P= W/t P=10300.5J/25.0s P=412.02w=4.1 x 10^2w The average power is 4.1 x 10^2w Power is the amount of work done over a period of time. Power is measured in Watts (W). It is the work/ time ratio, and the formula for power is P = W/t. Power is a time based quantity. It relates to work because, power is basically the rate at which work is done. This means work is essential for power to be present. This sample question asks to solve for the average power. To calculate the average power we must first find the total work done. Once we have both the time and the total work done, we can plug the values into our formula for power and end up with a answer of approximately 4.1*10^2 W. This sample question allows us to see how power is related to work and time, therefore, the I Can Statement and the question from the document are answered.

Physics 20 Unit C Assessment

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