Recent content by pharm89

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    Mechanical Energy in a Pendulum

    Homework Statement Hi,I am taking a grade 11 physics correspondance course and I have been stumped for the last few weeks on what they are asking in a particular question. A pendulum was set up and measurements were made to enable the mechanical energy to be calculated at the start...
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    Conservation of energy of bullet into sandbag

    Thanks..I guess I need to break down the question into different parts. -exploring the law of conservation of energy This law states that the total amount of energy in the universe remains constant. discover all forms of energy in the question: -Thermal energy - converted in the sandbag...
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    Conservation of energy of bullet into sandbag

    Thanks for the help. The question makes much more sense now. There are two other parts to that question which are theory based. They are: b) Explain 1 application of the first law of thermodynamics in this example. My answer: Energy both enters and leaves this system. The thermal...
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    Conservation of energy of bullet into sandbag

    True, of course that would not be right:))) Thanks for the tips. The collision of the bullet and the sandbag 1/2(0.015 kg) (840m/s)^2 = 1/2 (0.015kg) (705600) =5292 J The rising of the bullet and sandbag Emechanical = mgh =(0.015 kg (bullet) + 24 kg (sandbag))(9.8m/s^2)(0.8m)...
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    Conservation of energy of bullet into sandbag

    Thanks for the help. Am I understanding the question any better. I tried the problem again and did calculations for both parts. 1. the collision of the bullet and the sandbag Mechanical Energy - Ek + eg = Eh since h = 0 1/2 (0.015 kg) (840 m/s) =1/2 (12.6) = 6.3 J 2. The rising...
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    Conservation of energy of bullet into sandbag

    Homework Statement Hi I am taking a self study physics course and not sure about the following question. Any help would be much appreciated. Thanks 1. A 15.0 g bullet is fired at 840 m/s into a ballistic pendulum consisting of a 24.0 kg bag of sand. If the bag rises to a height of...
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    Kinetic and Potential Energy

    Thank you. E thermal = 67.6 J + 3600 J (amount of work done) =3667.6 J
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    Kinetic and Potential Energy

    Homework Statement Hi, I am taking a physics correspondace course and wondering if I am interpreting these questions correctly. Thanks Homework Equations 1. a) A force of 24 N is used to push a smooth box 3.6 m across a smooth floor. What is the final kinetic energy of the box? b)...
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    Work and transformation of energy

    Thanks for the help. therefore the conditions for work are met. g= 46kg X 9.8 N =450.8 N W=450.8 N X 6.2 m W=2794.96 J = 2795 J
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    Work and transformation of energy

    whoops question c and d (c) A 46 kg woman climbs a flight of stairs, 6.2 m high. How much work does she do. (d) A force of 104 N is applied to a 26 kg mass at an angle of 32 degrees to the horizontal. Calculate the work done when the mass is moved through a horizontal distance of 5.0m...
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    Work and transformation of energy

    Homework Statement Hi I am taking a grade 11 physics correspondance course and would like to know if iam on the right track. thanks in advance. 1. (a) A force of 280 N is required to push a box 8.2 m across a rough floor. How much work is done? (b) Explain 1 transformation of...
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    Help with interpreting lab assignment: factors affecting Friction

    thanks very much for your comments Pharm89
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    Help with interpreting lab assignment: factors affecting Friction

    Homework Statement I am responsible for coming up with an experiment to determine how the following 3 factors affect the force of friction: weight, surface area and speed. The directions given inlcude: "You will need to measure the magnitude of the force of friction. Since the...
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    Motion Near Earth's Surface

    Thanks for your help with this problem. It makes more sense to me know. Pharm89
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    Motion Near Earth's Surface

    i substituted 9.8 m/s^2 for F_net, still using the same varibles for mass and acceleration and solved for X. and Icame up with 0.51 N? Or is my method simply wrong. Thanks Pharm89
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