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Number of cores and sectional area of cable lgj-120, thank you
First of all, lgj-120 is not a cable. This is ACSR
Lgj-120 model core number is not necessarily the same
In general, the section of 26 aluminum 7 steel is 134.49
Equations a: B: C: D = 4:3:2:1 a-2b-3c + 4D = - 8 equations {3 divided by (2x + y) - x + 2Y = - 0.5 5 divided by 4x + 2Y + x-2y = - 21 / 2 help me
First:
From the formula, a = 4D, B = 3D, C = 2D
We can get d = 2, a = 8, B = 6, C = 4 when we bring in 2
Second: 5 divided by 4x + 2Y is reduced to 2.5 / 2x + y
Add Formula 1 and formula 2 to get 1 / 2x + y = - 2
So 2x + y = - 1 / 2
The title you gave is a little irregular
Can only do to this step, if there is any problem, the title photo sent to the above
It's quite clear
What is the law of voltage in series circuit?
Forget about it!
In series circuit,
U (total) = u (1) + U (2)
In parallel circuit,
U (total) = u (1) = u (2)
Is one plus one a fraction
Because: the basic concept of fraction is like a / b. A and B are integers. The integral with unknown number in B and B is not equal to 0 is called fraction. Where a is the numerator of fraction and B is the denominator of fraction
To judge whether a formula is a fraction, don't look at whether it is a / b form
(1) The denominator of a fraction must contain an unknown number
(2) The denominator value cannot be zero. If the denominator value is zero, then the fraction is meaningless
So: one plus one a is a fraction
Two lamps L1 "220 V 40 W" and L2 "220 V 60 W" are connected in series and then connected in a 220 V circuit to calculate the actual power of the two lamps
The change of filament resistance is not considered
R1=U1^2 /P1=(220V)^2/40W=1210Ω
R2=U2^2/P2=(220V)^2/60W=(2420/3)Ω
I=U/(R1+R2)=220V/[1210Ω+(2420/3)Ω]=(6/55)A
P real 1 = I ^ 2xr1 = [(6 / 55) a] ^ 2x1210 Ω = 14.4w
P real 2 = I ^ 2xr2 = [(6 / 55) a ^ 2x (2420 / 3) Ω = 9.6w
I think the process should be like this If it's wrong, please forgive me!
And I'm sorry, I can't play the division sign of the sum of squares So, it might be a bit of a hassle to watch
If a, B, C are all nonzero real numbers, and 2 ^ 6A = 3 ^ 3B = 6 ^ 2C, prove: 1 / A + 2 / b = 3 / C
Certificate:
Let 2 ^ 6A = 3 ^ 3B = 6 ^ 2C = k, then 6A = log (2, K) = 1 / log (k, 2)
3b=log(3,k）=1/log(k,3)
2c=log(6,k)=1/log(k,6)=1/[log(k,2)+log(k,3)]=1/[1/6a+1/3b]
That is, 1 / 2C = 1 / 6A + 1 / 3B leads to 1 / A + 2 / b = 3 / C
The key to solve this problem is to set K, then carry out the transformation of finger pairs, and then carry out the logarithmic operation
The electromotive force of the power supply is 1.5V, the internal resistance is 0.12ohm, and the external resistance is 1.38ohm
The formula and calculation process are more detailed
According to Ohm's law, the current I = ε / (R + R) = 1.5 / (1.38 + 0.12) = 1a
End pressure U = IR = 1 * 1.38 = 1.38V
The hall has eight similar cylindrical wooden columns, each of which is 5 meters high and 1.2 meters long at the bottom
If 0.8 kg of paint is used per square meter, how many kg of paint are needed to brush these wooden columns
36.8
The product of the height of 5 and the circumference of the bottom surface of 1.2 meters is the area of the cylinder, and then multiplied by 0.8 is the required paint
I hope I can help you~~~
As for the rated power of electrical appliances, the correct one in the following statement is ()
A. The more electric energy consumed, the greater the rated power. B. the smaller the current in the appliance, the smaller the rated power. C. when the voltage on the appliance decreases, the rated power remains unchanged. D. when the appliance does not work due to power failure, the rated power is zero
When an electrical appliance works normally, the voltage at both ends is the rated voltage, and the power under the rated voltage is the rated power. Therefore, a and rated power have nothing to do with the amount of electric energy consumed by the electrical appliance, so statement a is wrong, which is not in line with the meaning of the topic. B and rated power have nothing to do with the current in the electrical appliance, so statement B is wrong, which is not in line with the meaning of the topic. C. when the voltage on the electrical appliance decreases, it will be wrong D. when the electric appliance does not work due to power failure, the actual voltage is zero, the actual power is zero, and the rated power does not change, so the statement D is wrong, which does not conform to the meaning of the topic
What is [(a-b) & # 178; + 4 (a-b) + 4] / (a-b + 2) equal to,
The original formula = [(a-b) + 2] & # 178; / (a-b + 2)
＝A-B＋2
It's a square. Notice
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