Gravity potential energy E = MGH, unit: for example, how to see the installed capacity of hydropower? How to budget the daily power generation?

Gravity potential energy E = MGH, unit: for example, how to see the installed capacity of hydropower? How to budget the daily power generation?

This is a derived formula, that is, the gravitational potential energy of the object can be obtained by the mass m of the object, the gravitational acceleration g of the object's position, that is, the height h of the object relative to the reference plane. The mass is absolute, the international system unit is kg, and the acceleration value will be given in the general question, the international system unit is m / (s * s), the height is relative, and the international system unit is m

Fill in the number according to the rule: 2,5,8,11, what number should be filled after: a.12 b.13 c.15 d.17 e.18

I have answered this question before
In the first case:
5-2=3
8-5=3
11-8=3
The number forms the arithmetic sequence
So? = 11 + 3 = 14
2,5,8,11,14
The second situation
Because 2 + 5 = 7
5+8=13
8+11=19
11+x=y
The sum of two adjacent terms is a sequence of increasing prime numbers, and the interval between them is not counted
That is, 7, (11), 13, (17), 19, (23), 29
So y = 29
So x = 29-11 = 18
So the answer is f

What is 888888 * 888888 1 + 2 + 3 + 4 + 5 + 6 + 7 + 6 + 5 + 4 + 3 + 2 + 1?
Fractions, integers

888888×[(1+2+3+4+5+6+7+6+5+4+3+2+1)/888888]
=888888×(1/888888)×(1+2+3+4+5+6+7+6+5+4+3+2+1)
=(888888/888888)×(1+2+3+4+5+6+7+6+5+4+3+2+1)
=1+2+3+4+5+6+7+6+5+4+3+2+1
=2×(1+2+3+4+5+6)+7
=2×[(1+6)×6/2]+7
=(1+6)×6+7
=42+7
=49

What is the sum of all the digits in the final product of formula 2 × 3 × 5 × 7 × 11 × 13 × 17?

Remember that 7 × 11 × 13 = 1001
Then 2 × 5 = 10
The rest only need to calculate the sum of 3 × 17 digits
Because he is not more than 1000
So the final form must be abcabc
In fact, it's 3 times 17, and the sum of each digit is multiplied by 2
So 51 is 6 × 2 = 12

X + y = 3, y + Z = 4, Z + x = 5. There are several methods to solve the linear equation of three variables

X=2,Y=1,Z=3,
I feel that there is only one kind of elimination method. If there are other methods, they still need elimination in essence
Method 1: the second formula minus the first formula, get z-x = 1, and then add the third formula, get z = 3, and then substitute into the first formula and the second formula, get x = 2, y = 1,
Method 2: add the three formulas and divide them by 2 to get x + y + Z = 6, then subtract the three formulas to get x + y + Z = 6
X=2,Y=1,Z=3,
Method 3: from the first formula, get y = 3-x, substitute into the second formula, get z-x = 1, plus the third formula, get z = 3, substitute into the first formula and the second formula, get x = 2, y = 1,

How many tons is 195 kg?

1 ton = 1000 kg
therefore
195÷1000=39/200
therefore
195 kg is 39 / 200 tons

Given the function f (x-1) = x2-3, the value of F (2) is ()
A. -2B. 0C. 1D. 6

Let X-1 = 2, we can get x = 3, so f (2) = 32-3 = 6, so D

Solve the following equation (1) 18 = 5-x

18=5-x
x=5-18
x=-13

[2sin50º+(1+√3tan10º)cos10º]/√2cos5
So how did 2cos50 & ordm; come from

The original formula = (2sin50 & ordm; + cos10 & ordm; + √ 3sin10 & ordm;) / √ 2cos5 & ordm; = [2sin50 & ordm; + 2 (1 / 2 * cos10 & ordm; + √ 3 / 2sin10 & ordm;)] / √ 2cos5 & ordm; = (2sin50 & ordm; + 2cos50 & ordm;) / √ 2cos5 & ordm; = 2 √ 2 (√ 2 / 2sin50 & ordm; + √ 2 / 2cos

When x = - 3, the value of the algebraic formula ax5 + BX3 + cx-8 is 6. Try to find the value of ax5 + BX3 + cx-8 when x = 3

When x = - 3, the algebraic formula ax5 + BX3 + cx-8 = - 243a-27b-3c-8 = 6; when - 243a-27b-3c = 14, namely 243a + 27b + 3C = - 14, x = 3, ax5 + BX3 + cx-8 = 243a + 27b + 3c-8 = - 14-8 = - 22