Physics questions, a little difficult, physics questions not too difficult?

Updated on educate 2024-05-25
7 answers
  1. Anonymous users2024-02-11

    Regular is the left thing right code, and now the left side is heavier than the right side.

    For the first time, first stop the game code at 5G, put the crystal on the left, and balance it.

    The second time, the code is moved to 5G, the crystal is placed on the right, and the crystal is placed on the left to make the balance, and the balance is 5+.

    The third time, the code is moved to 5G, and the 7G crystal is placed on the right, and the crystal is placed on the left to make the balance, and the balance is 5+.

    Fourth, the code is moved to, and the 7g and 14g crystals are placed on the right, and the crystals are placed on the left to make the balance, and the balance is obtained.

    The sum of the four gains:

    over~~

    It's so tiring to watch, so let's give more points.

  2. Anonymous users2024-02-10

    The first time, the second time put the first one on the right disk as a weight, plus the game code can be called 7g, the third time to use one, the second time as a weight, plus the game code can be called a total of 14g, the fourth time to add the first three together as a weight, and then move the game code right 1g, so that you can weigh out, four times a total.

  3. Anonymous users2024-02-09

    vb=(xab+xbc) (2t).

    Grinding (2 ms.)

    m s so the boy c option is correct.

  4. Anonymous users2024-02-08

    A student wants to use a side-elimination battery to charge his Walkman No. 5 nickel-chromium battery, and there are strict requirements for the charging current during the charging process, please use the equipment provided in the picture (there is: a switch, 1 No. 5 nickel-chromium battery, an ammeter, a sliding rheostat, and a battery pack composed of 3 lead-acid batteries) to design a charging circuit, and the following questions:

    1.Tell us your opinion on the ammeter connection and why?

    2.What role do sliding rheostats play in a circuit? Asked by:

    Broken bridge remnants. Scholar.

    Level 2. 1. The battery and ammeter are connected in series.

    Slide rheostat and 1 in parallel.

    The battery is then connected in series with 1,2.

    The ammeter controls the amount of current passing through the battery, so that it is possible to know exactly how much power is flowing through the battery.

    The sliding rheostat shares the electrical backup current.

    Divide the voltage in series, divide the current in parallel, and how to control the current in series with all the required voltage.

  5. Anonymous users2024-02-07

    The actual division value of the thermometer = 100 (94-4) = (10 9) The actual temperature of water t = (22-4) * (10 9) = 20 Let the temperature sought be t

    t=(t-4)*(10/9)

    t=40℃

  6. Anonymous users2024-02-06

    =w useful w total = gh fs = g fn = g fn = g (g + g movable pulley) = g pants with g total.

    f=1 n·g total.

    g is the weight of the object, h is the height of the object rising, f is the size of the tensile force, s is the height of the rope envy pure fiber rises, and n is the number of rope windings (connected to the movable pulley).

    Or: =w useful w total = gh g total h = gh (g+go) h=g (g+go).

    G is the total weight of the moving pulley and the object, and GO is the weight of the imitation wheel of the moving skating brother.

  7. Anonymous users2024-02-05

    In the above question, the water in Xiaoxin's water tank is often not warm enough in winter, so the students installed a "220 V, 2000 W" electric heating pipe in the Qingshen sock water tank, and used a small "6 V, 1W" light bulb and a 10 m resistor.

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