How much a current can the national standard BVR six square millimeter copper core power line (actually) load?

How much a current can the national standard BVR six square millimeter copper core power line (actually) load?

The rated current in the conduit is 34 amperes and 40 amperes for a short time (one hour)

Given real numbers a, B, C, and a + B + absolute value [radical (C-1) - 1] = 4 radical (A-2) + 2 radical (B + 1) - 4, find the value of a + 2b-3c

A + B + | radical (C-1) - 1 | = 4 radical (A-2) + 2 radical (B + 1) - 4
(A-2) - 4 radical (A-2) + 4 + (B + 1) - 2 radical (B + 1) + 1 + | radical (C-1) - 1 | = 0
[radical (A-2) - 2] & sup2; + [radical (B + 1) - 1] & sup2; + | radical (C-1) - 1 | = 0
Because every term on the left side of the equation is nonnegative, make the result equal to 0
It must have root (A-2) - 2 = 0, root (B + 1) - 1 = 0, root (C-1) - 1 = 0
The solution is a = 6, B = 0, C = 2
a+2b-3c=6+2×0-3×2=0

Electric power and energy of refrigerator fan

TV (21 inch) CRT 80-90w
Refrigerator 200L (compressor) 145-175w
Fan 400mm 50-60w
This is the general power consumption. What do you mean by electric energy?

The application of arithmetic mean and geometric mean
1. Find the maximum value of the function y = x (a-2x) (x > 0, a is a constant greater than 2x)
2. Try to find the minimum value of sin2 α · Cos2 α + (1 / (sin2 α · Cos2 α)) (2 means square)

Y = x (a-2x) = (1 / 2) * 2x (a-2x) = 2, if and only if (sin2 α · Cos2 α) ^ 2 = 1

How is the conversion between kilojoule and large calorie?
Can you give me an example? I'm not good at math, a little stupid Ha ha, for example, how to convert 502 kJ into kcal? Multiplication or division?

Calorie is the unit of calculation of heat. 1 calorie is the amount of heat required to raise the temperature of 1 gram of water by 1 degree centigrade after the food we eat is burned. Kilocalorie is the amount of heat required to raise the temperature of 1 liter of water by 1 degree centigrade. Kilocalorie is also called large calorie

A & # 178; - AB + B / A & # 179; + B & # 179; approximately

a+b

Two resistors are connected in series, R1 = 8 Ω, R2 = 15 Ω, the electric power of R1 is 2W, then the current passing through R1 and R2 is (), the voltage on R1 is () V, the voltage on R2 is () V, the electric power is () v

If two resistors are connected in series, R1 = 8 Ω, R2 = 15 Ω, and the electric power of R1 is 2W, the current passing through R1 and R2 is (0.5A), the voltage on R1 is (4) V, the voltage on R2 is (7.5) V, and the electric power of R2 is 3.75w

When an object moves in a straight line with uniform acceleration, the displacement in the fifth second is 10 m, and the displacement in the seventh second is 20 m

(1) V1 = 10m / S (the fifth second)
V2 = 20m / S (the seventh second)
a=(V2-V1)/t=(20-10）/(7-5)=5
(2)x=V0t+(at^2)/2
10 = VO + A / 2 (the fifth second)
20 = (VO + 2a) + A / 2 (the seventh second)
a=5

Calculation of resistance by electric power formula

P = I & # 178; R or P = u & # 178 / / R

As shown in the figure, a uniform thin rod with a mass of M and a length of L, with o point as the axis of rotation, swings freely from rest at an angle of θ with the vertical direction to the vertical position,
The angular velocity of the uniform rod m and the velocity of the object m after collision are calculated