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I don't know why,,, the electric field distribution of the same amount of heterogeneous charge that I have seen in high school is not perpendicular to the middle perpendicular line.
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Swimming upstream, the velocity of the boat against the rock is: (3-2)=2 (meters and seconds).
One minute later the boat is upstream of the bridge: 2*60=120 (m).
After the barrel falls into the water, it floats downstream and after one minute is downstream of the bridge: 1*60=60 (m).
Immediately after the discovery, turn the bow of the ship, at this time the barrel is 120 + 60 = 180 (m) When the ship goes down the water, the rate of the opposite shore is: 3 + 1 = 4 (m seconds) Set the barrel and drift x m to catch up, then:
180+x) 4 = x 1 to get 3x=180; x=60 (m) at this time, the barrel distance from the bridge: 60 + 60 = 120 (m).
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The pressure on the liquid surface at the upper end of the DC-shaped tube is smaller than that at the lower end, which is small: the pressure generated by the difference between the water column at the upper and lower ends.
The pressure at the upper end is equal to the atmospheric pressure, and the lower pressure is greater than the atmospheric pressure, so water flows out from the lower end.
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Solution: The force analysis of the box is shown in the figure above, in which ma is the inertia force generated by the acceleration of the car, 1) the friction force f can not provide the inertia force required for the accelerated movement of the box ma when the box slides.
f=mgμma>mgμ
a>gμ
2) If the box is turned over, it can be sure that it will be turned around the back and bottom of the box. The reason for turning over should be that the resultant moment around the lower edge of the back is not zero.
mah-mgl>0
a>lg/h
3) According to the correlation with l h, you can judge whether the box slides or overturns first.
If when AWhen g When A>lg H, the box slides while tipping over if >l h, then:
It can be judged that the box was tipped over first.
When A is LG H when A > g, the box overturns and the box slides at the same time.
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No matter how you wiring it, there are only two types of resistance: 2r 3 and 4r 3 (also the maximum resistance).
Even the 12, 13, 23, 24, 34, 16, 15, 65, 64, 45 post resistors are 2r 3;
Even the 25, 26, 35, 36 post resistors are 4r 3.
This kind of question is actually a test of your ability to simplify the circuit, once the circuit is simplified (like my diagram), then the rest of the calculation is very simple. Therefore, to do this kind of problem, the first thing is to simplify the circuit diagram and draw the equivalent circuit diagram.
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If you take two vertices, such as 1 and 6, then the resistor has a resistance value, and the total number of different resistance values (excluding 0 ohms) that can be obtained with this resistor is 2
The maximum resistance values are 4r 3 respectively
These 2 connections are 1 (4) to the other point where the resistance is 2r or between is 4r 3
To explain hi me in detail.
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The total number of different resistance values (excluding 0 ohms) that can be obtained with this resistor is 2, and the maximum resistance value is 4r 3, respectively
These 2 connections are 1 (4) to the other point where the resistance is 2r or between is 4r 3
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If you find two vertices such as 1 and 6, then the resistor has a resistance value, and the two different points correspond to a value, which may be the same or different, and it is calculated by connecting them in series and parallel. Then count them one by one.
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Classification discussion, the binding post is the boundary line, the same side (including no matter how the resistance value is one, there are 1 and 2 binding posts on different sides, and in each case, the circuit diagram can be simplified to calculate the resistance value.
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