Super easy physics questions, but you need to know Japanese

Updated on educate 2024-04-11
5 answers
  1. Anonymous users2024-02-07

    When the groundwater level drops by 3 m due to groundwater extraction, how does the pressure at point A, which is 10 m deep, change compared to before the drop? Find the division ratio. Select the corresponding value from the following quantities.

    Wet unit volume weight 18 3kn m3

    Saturated unit volume weight 19 6kn m3

    Unit volume weight in water 9 8kn m3

    Hope it helps

    kn m3 i.e. kilonewn per cubic meter.

    Therefore, kn m3 cannot be called a unit of density, only a density of gravity.

    Since g=mg, the essential meaning of these two is still the same, but it needs to be converted.

  2. Anonymous users2024-02-06

    When pumping water, the water table drops by 3 meters, and at a point of pressure it goes 10 meters underground, how much do you ask for a change. However, you can use the following reply to value all the works that come out.

  3. Anonymous users2024-02-05

    When the groundwater level drops by 3 m due to groundwater extraction, how does the pressure at point A, which is 10 m underground, change compared to the previous drop? Find that value. Just select the appropriate value from the individual quantities below.

    Wet unit volume weight 18 3kn m3

    Saturated unit volume weight 19 6kn m3

    Unit volume weight in water 9 8kn m3

  4. Anonymous users2024-02-04

    When the groundwater level is lowered by 3 m by drawing groundwater, what is the change in the pressure at point A at 10 m underground? However, please use the standard values provided below to answer.

    The wet weight of the soil is 18 3kn m3

    Dry weight of soil =

    Severe in the water (this is not very certain) 9 8kn m3

  5. Anonymous users2024-02-03

    Answer: Let the direction of the traction force f be positive.

    When there is a traction force f, f 1 = f - ff = 180n - 150n = 30n The acceleration of the object a = f 1 m = 30n 30kg = 1m s 2 The final velocity v of the object after 5 s of motion.

    In the absence of traction f, the acceleration of the object a2 = f = -150n = -150n 30kg = -5m s 2

    The time taken for the object to move at rest is t=(0-v) a2=1s, the distance traveled by the object in the whole process, s=average speed * total time = 1 2 v end * 5s + t) = 1 2 * 5 * 6 = 15 meters.

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