Given that I1 = 14.1sin (628t-50 °) a, I2 = 70.7sin (628t + 40 °) a, I = I1 + I2, find the maximum and effective value of I,

Given that I1 = 14.1sin (628t-50 °) a, I2 = 70.7sin (628t + 40 °) a, I = I1 + I2, find the maximum and effective value of I,

I1 = 10 * (2) ^ (1 / 2) sin (628t-50 °) a, I2 = 50 * (2) ^ (sin (628t + 40 °), the difference between the two vectors is exactly 90 degrees, so the maximum value of I is (200 + 5000) ^ (1 / 2) = 72.11a, and the effective value is 72.11 / (2) ^ (1 / 2) = 50.998a

In the parallel circuit, the current of each branch is I1 = 60sin (26t + 135 degrees), I2 = 20sin (26t-45 degrees), and the total current I is calculated

In parallel circuit, the current of each branch is I1 = 60sin (26t + 135 degrees), I2 = 20sin (26t-45 degrees)
Parallel circuit, the voltage is the same, the current is equal to the sum of the branch currents, I total = I1 + I2 = 60sin (26t + 135 degrees) + 20sin (26t-45 degrees) = 80sin (26t + 135 degrees - 45 degrees) = 80sin (26t + 90 degrees)

If I = I1 + I2, I1 = 10sin ω TA, I2 = 10sin ω a, then the effective value of I is

AC instantaneous value:
u = Um * sin(φ+ωt)
i = Im * sin(φ+ωt)
Effective value I = im / √ 2 = 0.707 * im
I1 = 10sin ω TA, I2 = 10sin ω a (less t), I1 equals I2:
I = I1 + I2 = 0.707 * 10 * 2 = 14.14 A

The weight ratio of goods in warehouse A and warehouse B is 7:8. If 6 tons are transported from warehouse B to warehouse A, the goods in the two warehouses will be equal. How many tons of original goods are there in warehouse A and warehouse B?

(6 × 2) △ 87 + 8-77 + 8, = 12 △ 115, = 180 (tons); a: 180 × 77 + 8 = 84 (tons); B: 180 × 87 + 8 = 96 (tons); a: the original goods in warehouse A are 84 tons, and the original goods in warehouse B are 96 tons

(x2 + 7x + 10) / (x + 1) and x > - 1 to find the minimum value of the equation
Sorry, it's not an equation. It's the minimum of an equation~

First of all, the denominator x2 + 7x + 10 can be sorted into x (x + 1) + 6 (x + 1) + 4, so dividing by the denominator x + 1, we get (x + 1) + 4 / (x + 1) + 5. Because x > - 1, so x + 1 > 0, then according to the basic inequality, x + 1 + 4 / (x + 1) is greater than or equal to 4, so the original formula is greater than or equal to 9
That is, the minimum value is 9

Three mathematical problems on the first day of junior high school
1. For a two digit number, the number of one digit is 2 larger than the number of ten digits. If the number on the two digits is transposed, the sum of the new two digits and the original two digits is 44, and the original two digits are obtained
2. For a three digit number, the sum of the three digits is 15, the ten digit number is 5 less than the hundred digit number, and the ten digit number is 1 / 3 of the one digit number
3. There is a column of numbers, which are arranged into 1, - 3,9, - 27,81, - 243 according to certain rules
The sum of some three adjacent numbers is - 1701. What are the three numbers?
It's mainly about ideas

Let the number of digits be x, then x + (X-2) * 10 + 10x + X-2 = 44, x = 3, primitive 13
Let ten digits be x, then x + 5 + X + 3x = 15, x = 2, original number 726
Let the first be x, then x + (- 3) x + 9x = - 1701, x = 243 243 - 729 2187

If Sn = 36, s2n = 54, then s3n =?
My personal understanding is that Sn, s2n and s3n are a set of equal ratio sequences,
Therefore, s2n / Sn = s3n / s2n
So we get: 54 / 36 = s3n / 54
Finally: s3n = 81, but the correct answer is 63? I don't understand

Method 1:
Sn, s2n Sn, s3n-s2n are in equal proportion sequence
∴(S2n-Sn)/Sn=(S3n-S2n)/(S2n-Sn)
∴S3n=63
Method 2: because this problem has nothing to do with N, let n = 1
If A1 = S1 = 36, A2 = s2-s1 = 54-36 = 18, the common ratio is obviously 1 / 2, A3 = 9
So S3 = a1 + A2 + a3 = 63

1. The number a is two fifths of the sum of the two numbers B and C, the number B is five ninths of the sum of the two numbers a and C, and the number C is 25
2. A car from a to B, first one-third of the whole journey, then three fifths of the whole journey, at this time more than 14 kilometers, find out the distance between a and B how many kilometers
3. Someone added a batch of parts, processed one fifth of all the parts in the first time, and made eight more parts in the second time than in the first time. At this time, the processed parts are two less than those not processed. How many parts are there in this batch
4. From land a to land B, two fifths of them are uphill roads, one fourth are flat roads, and the rest are downhill roads. A bicycle goes back and forth between land a and land B for a total of 3 kilometers uphill. How many kilometers is the distance between land a and land B

Let a be x and B be y
Set up equations according to the meaning of the question
x=2/5(y+25)
y=5/9(x+25)
The solution is x = 20, y = 25
So a + B + C = 20 + 25 + 25 = 70
The distance is x km
Equations: X / 3 + 3x / 5-x / 2 = 14
X = 420 / 13
Suppose there are x parts in this batch
Equations: 2 (2x / 5 + 8) + 2 = x
We get x = 90
The distance between a and B is x km
Equations: 2x / 5 + (x-2x / 5-x / 4) = 3
We get x = 4
OK

The cross section of a reservoir dam is trapezoidal ABCD, and the crest width CD = 3M,
The slope ad = 16m, the dam height is 8m, the slope BC is 1:3, and the slope angle A and the dam bottom width ab of the slope AD are calculated in detail

CE ⊥ AB at point C and DF ⊥ AB at point d
∵CE⊥AB,DF⊥AB,CD∥AB
｜ rectangular cdfe
∵ dam height 8, CD = 3
∴CE＝DF＝8,EF＝CD＝3
∵AD＝16
∴sin A＝DF/AB＝8/16＝1/2
∴∠A＝30
∴cosA＝√（1- sin²A）＝√（1-1/4）＝√3/2
∴AF＝AD* cosA＝16*√3/2＝8√3
∵ the slope BC is 1:3
∴BE＝CE*1/3＝8/3
Ψ AB = AF + EF + be = 8 √ 3 + 3 + 8 / 3 = 8 √ 3 + 17 / 3 (m)

Fill in the operation symbols between the numbers to make the equation hold: 2 = 1; 9 = 73

2×2-2-2÷2=1
9×9-9+9÷9 =73