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    Lower Secondary Science

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

      superkiasume:
      Hi, can someone help my son with the following questions:


      (a) The engine of a car of mass 800kg can develop a constant 80kW of power. If its maximum speed on a flat road is 100km/h, determine the resistance to its motion.

      (b) With the engine working at the same rate, find the acceleration of the car when is travelling at a speed of 50km/h, if the resistance is constant.

      Thank you.
      For Part a, you can use the formula that at constant velocity

      Power = Force X velocity

      100km/hr converts to 27.8 m/s

      80,000 W = Force X 27.8m/s

      So Force = 2880 Newtons.

      The driving force pushing the car forward is equal to the force of friction because it is moving at constant velocity.

      Usually students remember that Power = Energy / time
      At constant velocity the work done to push the car is the force times the distance. Then if we divide by time we get power. So if the car travels 27.8 meters in one second, then force X 27.8m is the work - then divide by 1 second of time. So this is equivalent to saying that Power = Force X velocity when moving at constant velocity.

      part (b) is a bit confusing to me. If the engine is working at the same rate, that means it is delivering the same power. But the statement that the resistance is constant is confusing. Most of the resistance to a car is air resistance which depends on speed. So it is a bit hard to decipher part (b).

      1 Reply Last reply Reply Quote 0
      • D Offline
        Dr.033430Daniel
        last edited by

        spicy:
        Hi

        I'm a stranger to Physics :scratchhead: , do appreciate help with this question pl. Thanks :please: 😢

        http://i60.tinypic.com/xp90k3.jpg\">
        The best place to start is

        Volts = Work/charge

        I've got the charge - they gave me just the charge of 1 electron, so calculating current in the circuit is not needed.

        The key skill is to be able to look at that circuit diagram and realize that the 10 Ohm resistor has 8 Volts across it.

        The thought process to arrive at that goes like this:

        The two 10 Ohm resistors in parallel have an equivalent resistance of 5 Ohms. So this means there is a 5 Ohm and 10 Ohm resistor in series. For resistors in series the voltage across each resistor is determined by the proportion of the total resistance. This means that the 12 Volts of the battery is divided up into three parts. 2/3 of the voltage is dropped across the 10 Ohm resistor and 1/3 across the 5 Ohm resistor since they are in series. If you can \"see\" that, you just write

        8Volts = Work/ 1.6E-19 C and you are done. The answer is B.

        Being able to recognize how voltages are distributed across resistors in series is an important skill for O'Level pure physics and beyond. If you had 12 volts across two 10 Ohm resistors in series, each has 6 Volts. If you had 12 volts across a 1 Ohm and 11 Ohm resistor in series, the 1 Ohm resistor has 1 Volt across it and the 11 Ohm resistor has 11 Volts. The proportion of the total resistance is the proportion of the voltage. This is true for resistors in series, but not in parallel.

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        • D Offline
          Dr.033430Daniel
          last edited by

          sunnymoon:
          Can someone help me with this question?

          A basket is being carried by a boy at a constant height above the ground. Is work being done?
          Thanks for your help
          The work done is force times distance traveled in the same direction as the force. The boy will have to exert an upward force on the basket to hold it up, but if it is at constant height there is no distance traveled in the direction of the upward force, so therefore no work done.

          In the horizontal direction, the general assumption in these kinds of questions is that there is no horizontal force needed if a person is walking along at constant velocity and air resistance to the basket is negligible (since it is at a slow speed). So the assumption is that no work is needed to move it horizontally at constant velocity with no friction.

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          • S Offline
            spicy
            last edited by

            :thankyou:


            Your help is much appreciated :lovesite:

            1 Reply Last reply Reply Quote 0
            • S Offline
              superkiasume
              last edited by

              Dr.Daniel:
              superkiasume:

              Hi, can someone help my son with the following questions:


              (a) The engine of a car of mass 800kg can develop a constant 80kW of power. If its maximum speed on a flat road is 100km/h, determine the resistance to its motion.

              (b) With the engine working at the same rate, find the acceleration of the car when is travelling at a speed of 50km/h, if the resistance is constant.

              Thank you.

              For Part a, you can use the formula that at constant velocity

              Power = Force X velocity

              100km/hr converts to 27.8 m/s

              80,000 W = Force X 27.8m/s

              So Force = 2880 Newtons.

              The driving force pushing the car forward is equal to the force of friction because it is moving at constant velocity.

              Usually students remember that Power = Energy / time
              At constant velocity the work done to push the car is the force times the distance. Then if we divide by time we get power. So if the car travels 27.8 meters in one second, then force X 27.8m is the work - then divide by 1 second of time. So this is equivalent to saying that Power = Force X velocity when moving at constant velocity.

              part (b) is a bit confusing to me. If the engine is working at the same rate, that means it is delivering the same power. But the statement that the resistance is constant is confusing. Most of the resistance to a car is air resistance which depends on speed. So it is a bit hard to decipher part (b).

              Thank you, Dr Daniel šŸ™‚

              1 Reply Last reply Reply Quote 0
              • B Offline
                Bookshop
                last edited by

                Bookshop:
                Bookshop:

                The differences in the yearly TYS are:

                1) The answer key provided. In terms of the answers provided, SAP seems to have come a long way. The answer keys used to be littered with errors here and there. But they have improved signficantly, and now generally provide answers that are more trusted than the other publishers.
                2) Pricing. Popular bookshops are offering higher discounts for EPH, as it's their in-house brand.
                3) Size of the book. Shinglee and EPH books are of A4 sizes, which is the O-Level standard. Also, their pages have been specially designed so that they can be torn out from the book for separate yearly revision.

                You have to decide which of the above factors is the most important.
                [quote=\"hopefulparent\"]I am thinking of buying the yearly TYS for my boy. He is taking the O-levels in a few months' time.

                Does anyone have any experience as to which TYS is preferable? I saw EPH, SAP, Shinglee and some other brands at Popular. Are they all the same?

                Thank you.

                Thank you for the advise, Bookshop.

                Do you have any recommendations for Physics and Chemistry revision books, where there are notes and formulas. Do you think they will be useful for last minute revision?[/quote]
                EPH Science At Your Fingertips - pocket-size notebook with summarised notes.

                Smashing Exams Physics and Chemistry - complete guidebook with full notes.

                1 Reply Last reply Reply Quote 0
                • T Offline
                  tutor76
                  last edited by

                  Its time to try another new question for budding tutor out there šŸ™‚


                  http://i60.tinypic.com/2sbx1j4.jpg\">

                  1 Reply Last reply Reply Quote 0
                  • S Offline
                    sunnymoon
                    last edited by

                    Can someone help me with this question:

                    Suggest reasons why copper pots are preferable to aluminium pot for cooking acidic food?

                    1 Reply Last reply Reply Quote 0
                    • T Offline
                      tutor76
                      last edited by

                      Copper is lower in the metal reactivity series ,therefore it does not react with acid ,aluminium is higher in the metal reactivity series ,will react to form aluminium salt ,bad for human health .

                      1 Reply Last reply Reply Quote 0
                      • T Offline
                        tutor76
                        last edited by

                        Another set of comb chemistry for student to

                        try ! http://i60.tinypic.com/ehx06h.jpg\">

                        1 Reply Last reply Reply Quote 0

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