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    O-Level Physics

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • D Offline
      Dr.033430Daniel
      last edited by

      Kafer:
      May i know how to solve this qn?


      an unknown translucent liquid has a reflective index of 1.75

      what is the maximum angle of refraction within the liquid if a neam of light passes fr air into the liquid?
      A. 32.7
      b. 34.8
      c. 55.2
      http://i59.tinypic.com/1072wt5.jpg\">

      I recommend in my O'Level classes that there is a single formula and procedure that can be used to solve all refraction problems at O'Level. The first step is to draw a diagram of the problem and clearly label the incident and refracted angles and indices of refraction. Next, use the Snell's Law equation and plug in the appropriate information to solve.

      In this particular case it is asking for the maximum angle of refraction inside the liquid. The maximum angle of refraction occurs when the incident angle is maximum, meaning the incident angle is 90 degrees. Just plugging in the two indices of refraction and the 90 degree angle into the formula gives you the angle of refraction. Both the incident and refracted angle are measured from the normal.

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      • K Offline
        Kafer
        last edited by

        Thanks! Dr Daniel.


        I hv the following question.

        1. If a basketball player can jump 2m into the car, calculate the velocity he has when he just leaves the ground. Take acceleration due to gravity to be 10m/s

        2. A person standing in front of a smooth vertical wall hits 2 blocks of wood with a frequency of 0.8 hz. What is the shortest distance he should stand fr the wall so that he will not detrct any echo? Take the speed of sound in air as 320m/s.

        3. A tuning fork is an instrument used to tune musical instrument. A note is produced by the tuning fork and the distance between the adjacent rarefraction and compression is 1.5m. What is the frequency of the tuning fork if the speed of sound is 300m/s?

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        • D Offline
          Dr.033430Daniel
          last edited by

          Kafer:
          Thanks! Dr Daniel.


          I hv the following question.

          1. If a basketball player can jump 2m into the car, calculate the velocity he has when he just leaves the ground. Take acceleration due to gravity to be 10m/s2
          I interpret this question to mean that a basketball player jumps a vertical distance of 2m, so we are finding his initial velocity. For students in our O'Level prep classes, there are 4 equations of kinematics that I recommend they have memorized. Then we plug the numbers in to the one that fits. This is the fastest way to do these kinds of problems. The solution below starts with one of the 4 equations of kinematics.

          v^2 = u^2 + 2ad
          0 = u^2 + 2 (-10m/s2) (2m)
          u = 6.32 m/s

          I called the upward direction positive, so the initial velocity u is + and d is + (jumping up)
          The downward direction is negative, so a is - (gravity acting downward)
          the final velocity v is 0 when the jumper reaches the 2m height.

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          • D Offline
            Dr.033430Daniel
            last edited by

            Kafer:
            Thanks! Dr Daniel.


            I hv the following question.

            2. A person standing in front of a smooth vertical wall hits 2 blocks of wood together with a frequency of 0.8 hz. What is the shortest distance he should stand fr the wall so that he will not detrct any echo? Take the speed of sound in air as 320m/s.
            I added the word \"together\" in the question for clarity. A person is banging two blocks of wood together, similar to clapping hands but instead he is clapping with wood so it is loud. The sound from this clapping goes out to the wall and is reflected back. The idea is that if he stands a certain distance from the wall, the echo will arrive just as he is clapping again. This way he does not hear the echo.

            So first we need to find the time between claps. The period = 1/frequency, so the time between claps is 1/0.8Hz = 1.25 seconds. In this amount of time the sound has to go to the wall and come back. So it takes half this time, or 0.625 seconds, to make it to the wall.

            Now we need to know the distance sound travels in 0.625 seconds. This will be the distance to the wall.

            0.625 s (320m/s) = 200m

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            • ChiefKiasuC Offline
              ChiefKiasu
              last edited by

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              • S Offline
                sc_gal
                last edited by

                Hi all! Am selling the following resources for Pure Physics:

                - Longman Effective Guide to O Level Physics
                - Longman O level Physics Workout (MCQ)
                - Perfect Guide: O level Physics Notes SOLD
                - GCE O level Topical Physics Guide (with detailed explanation)
                - Info Booklets (topical) from SCGS

                PM me if you're interested! šŸ™‚

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                • K Offline
                  kychua
                  last edited by

                  Gonna follow this thread šŸ™‚

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                  • S Offline
                    shurley197323
                    last edited by

                    Can someone help this question pls?


                    http://i62.tinypic.com/ou4cba.jpg\">

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                    • S Offline
                      shurley197323
                      last edited by

                      Please help!!!


                      A lawn roller has a diameter of 1.20 m and weighs 560N.

                      a) What is the minimum force needed to push the roller over a brick of height 0.100 m if the force is applied at the centre of the roller?

                      b) What is the minimum force needed to push the roller over a brick of height 0.100 m if the force is applied at the top of the roller?

                      Thank you ā¤

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                      • D Offline
                        Dr.033430Daniel
                        last edited by

                        shurley197323:
                        Can someone help this question pls?


                        http://i62.tinypic.com/ou4cba.jpg\">
                        The key to this problem is to understand that the weight of the beam is supported by the vertical component of the tension only. That vertical component is 4/5 times the tension of 1.20kN which equals 0.96 kN.

                        The force that the hinge exerts on the beam is the reaction force to the tension in the cable pulling horizontally. The tension pulls to the left horizontally, so the force that the question asks for is directed to the right. This is a 3-4-5 triangle, so this force is 3/5 times 1.20kN which equals 0.72kN.

                        The horizontal and vertical components of the tension can be thought of separately and they do not influence one another.

                        This question represents an important concept tested at A'Level and also in upper secondary in many IP schools. I've seen a few schools give this type of question for express O'Level physics also. It seems to be popular this year and is making the rounds in many schools.

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