The words "220 V, 5A, 3000 R / kW. H" are marked on a household electric energy meter, which is connected with a bulb of "220 V, 100 W", The meter turns 45 times in 10 minutes to calculate the energy consumed by the bulb in 10 minutes

The words "220 V, 5A, 3000 R / kW. H" are marked on a household electric energy meter, which is connected with a bulb of "220 V, 100 W", The meter turns 45 times in 10 minutes to calculate the energy consumed by the bulb in 10 minutes

The electric energy consumed by the bulb in 10 minutes w = 45r / (3000r / kW. H) = 0.015kwh = 54000j
Actual power of bulb p '= w / T = 54000j / 600s = 90W
The actual power of the bulb: P = 45 * 1000 * 60 / 93000 * 100 = 90W
The electric energy consumed by the bulb in 10 minutes is: w = 90 * 10 / (1000 * 60) = 0.015 kwh
According to the above conditions, this watt hour meter is slow
1 power P = 3600 × 45 / 600 × 3000 = 0.09kw. 2. Power consumption w = Pt = 90W × 600s = 54000j.
45%3000=0.015KW.h
An electric energy meter is connected to the home circuit, and the dial is marked with the words of 220V 5A and 3000r / kW. H
When only one electric appliance in the circuit works, it is observed that the electric energy meter turns 15 turns in 1 minute, then what is the electric energy consumed by using the electric appliance instead
3000 rpm is only 1 kwh, 1 rpm is that, 15 rpm is 0.005 kwh, equivalent to 0.005 kwh;
The equation is the fastest, 3000 / 1kwh = 15 / X; X = 15 / 3000 = 0.005; X = 15 / 3000 = 0.005;
A watt hour meter is marked with the words "220V 10A" and "3000r / kWh", indicating how much total power can this watt hour meter access at most?
Such as the title. Third day electricity convenient question
P=IU=220V×10A=2200W
Up to 2200j of work can be input in 1s
If the electric energy meter is marked with "220V 10A", how many lights of "220V 40W" can be connected to the circuit where the electric energy meter is located?
If these lamps work normally for one month (based on 30 days) with an average of 4 hours per day, how many kilowatts of electricity will they consume?
Rated current of each bulb (P = 0.220 a)
According to the "220 V 10A" of the electric energy meter, the current that the electric energy meter can load is 10 A
Then the light bulb connected with 220 V 40 W 0.18 a can be connected with 10 / 0.18 = 55.6
So it can connect up to 55 "220 V 40 W" lamps
1kwh = 1 kWh
The power of each lamp is 40W = 0.04kw
The power of 55 lamps is 55 * 0.04 = 2.2 (kw)
Power consumption of working 4 hours a day w = Pt = 2.2 * 4 = 8.8 (KWH)
If a month is calculated as 30 days, the total power consumption w = 8.8 * 30 = 264 (KWH) = 264 kwh
These lamps work normally for one month (30 days) with an average of 4 hours per day, consuming 264 kwh
To prove: 1-2sin α cos α / cos & # 178; α - Sin & # 178; α = Tan (π / 4 - α)
It is proved that the left = (Sin & # 178; α - 2Sin α cos α + cos & # 178; α) / (COS & # 178; α - Sin & # 178; α)
=(cosα - sinα)²/[(cosα - sinα)(cosα + sinα)]
=(cosα - sinα)/(cosα + sinα)
=(1-tanα)/(1+tanα)
=[tan(π/4) - tanα]/[1+ tan(π/4)*tanα]
=tan(π/4 - α)
=Right
So the equation is proved!
Let a = {(x, y) ly = x + 1, X ∈ r}, B = {(x, y) ly = - x2 + 2x + 3 / 4, X ∈ r}, find a ∩ B
Simultaneous y = x + 1, y = - x ^ 2 + 2x + 3 / 4
We get x + 1 = - x ^ 2 + 2x + 3 / 4, that is, x ^ 2-x + 1 / 4 = 0, that is, (x-1 / 2) ^ 2 = 0, x = 1 / 2
Then y = x + 1 = 3 / 2
So a ∩ B = (1 / 2,3 / 2)
If you make two equations equal, you can solve them,
And then we can get y
A and B are all point sets, so we can regard the following formula as two functions
The intersection of simultaneous functions is the solution of a ∩ B
I.e. y = x + 1 and y = - X & # 178; + 2x + 3 / 4, we obtain the equation 4x & # 178; - 4x + 1 = 0, i.e. (2x-1) &# 178; = 0
The solution is x = 1 / 2
Take it into y = x + 1 to get y = 3 / 2
So a ∩ B = {(1 / 2,3 / 2)}
It is known that X and y are positive numbers, and 2x + 5Y = 20. If the inequality μ≥ lgx + lgY holds, then the range of μ is
From X and y are positive numbers, and 2x + 5Y = 20, we get (2x + 5Y) * (2x + 5Y) = 400 ≥ 40xy, and 0 < XY ≤ 10,
Then lgx + lgY = LG (XY) ≤ 1, if μ≥ lgx + lgY is constant, then μ≥ 1
First, draw the feasible region of X and y, that is, the graph enclosed by X axis, Y axis and 2x + 5Y = 20, and then change the U ＞ = lgx + lgY to u > = lgxy. Let xy = k, then y = K / X draw its image and translate it to be tangent to 2x + 5Y = 20, then the intersection point is the required (x, y). Below, find y '= - K / X2, let y' = - 2 / 5, then K / x2 = 2 / 5, that is k = 2 / 5x2 (0
In the sequence {an}, A1 = 1sn is the sum of the first n terms of {an}. When n is greater than or equal to 2, Sn = an [1-2 / Sn] proves that {1 / Sn} is an arithmetic sequence
The second question is TN = S1 × S2 + S2 × S3 +. + Sn × s (n + 1)
1 / Sn = (n + 1) / 2, then Sn = 2 / (n + 1)
S1xS2+S2xS3+.+SnxS(n+1)=4(1/2 x 1/3 +1/3 x 1/4 +.)
=4(1/2 - 1/3 +1/3 -1/4 +.)
=4(1/2 -1/(n+2))
When n ≥ 2
an=Sn-Sn-1
Sn=（Sn-Sn-1）（1-2/Sn）
It is reduced to 1 / Sn = 1 / 2 + 1 / sn-1
So {1 / Sn} is an arithmetic sequence with tolerance of 1 / 2
The solution is Sn = 2 / (n + 1)
tn=2/2×2/3+2/3×2/4+．．．．．．2/n×2/（n+1）
∵1/n×1/（n+1） =1/n-1/（n+1）
Ψ TN = 2 / 2-2... Expansion
When n ≥ 2
an=Sn-Sn-1
Sn=（Sn-Sn-1）（1-2/Sn）
It is reduced to 1 / Sn = 1 / 2 + 1 / sn-1
So {1 / Sn} is an arithmetic sequence with tolerance of 1 / 2
The solution is Sn = 2 / (n + 1)
tn=2/2×2/3+2/3×2/4+．．．．．．2/n×2/（n+1）
∵1/n×1/（n+1） =1/n-1/（n+1）
∴tn=2/2-2/3+2/3-2/4+．．．．．．2/n-2/（n+1）
tn=1-2/（n+1）
TN = (n-1) / (n + 1) ask: I calculate 2 [1-2 / (n + 1)]
Given Sin & # 178; γ = Sin & # 178; α - sin α cos α Tan (α - β), prove Tan & # 178; γ = Tan α Tan β
In a hurry
∵sin²γ=sin²α-sinαcosαtan（α-β）①∴cos²γ=1-sin²γ=1-[sin²α-sinαcosαtan（α-β）】=1-sin²α+sinαcosαtan（α-β）cos²γ =cos²α+sinαcosαtan（α-β...
2. Let u = {(x, y) | x ∈ R, y ∈ r}, a = {(x, y) | 2x-y + m > 0}, B = {(x, y) | x + y-n ≤ 0}, then
What is the necessary and sufficient condition for P (2,3) ∈ a ∩ (cub)?
Because the set u = {(x, y) | x ∈ R, y ∈ r}, B = {(x, y) | x + y-n ≤ 0} then cub = {(x, y) | x + y-n ＞ 0} from P (2,3) ∈ a ∩ (cub), we can know that P (2,3) ∈ a, P (2,3) ∈ cub, that is, 2 * 2-3 + m ＞ 0,2 + 3-N ＞ 0, so m ＞ - 1, n ＜ 5, so the sufficient and necessary condition of P (2,3) ∈ a ∩ (cub) is m ＞ - 1, n ＜ 5