A cylindrical container with a bottom radius of 5cm is filled with water. Immerse a 157 cubic centimeter iron block in water. When the iron block is taken out, how many centimeters does the water surface drop?

A cylindrical container with a bottom radius of 5cm is filled with water. Immerse a 157 cubic centimeter iron block in water. When the iron block is taken out, how many centimeters does the water surface drop?


Water level drop = 157 ^ (3.14 × 5 & # 178;) = 2cm



An iron block is immersed in the water of a cylindrical glass container. The inner diameter of the bottom of the container is 20 cm. Take the iron block out of the water and drop it 4 cm. The volume of the iron block is


The density of water multiplied by (the area of the bottom of the container multiplied by the height of the water surface falling) ---- the mass of water (also the mass of iron block) is calculated
The mass of iron divided by the density of iron = the volume of iron



A 150 cubic centimeter iron block is completely immersed in a cylindrical container containing water. The diameter of the bottom surface of the cylindrical container is 10 cm. What is the height of the water surface rising? (the number is an integer.)
To calculate the process, learn better
Cubic energy and square division? I feel dizzy. Brother, your primary school is not as good as mine


The diameter of the bottom surface is 10 cm, so the area is 25 π
150 / 25 π is what we need
Of course --
If I tell you the volume of a cuboid, and then I tell you the length and width, how can you find the height?



The displacement of an object is 1 m in the first second and 2 m in the second If the internal displacement of NS is nm, then ()
A. The initial velocity of the object is zero. B. the acceleration of the object is 1m / S2C. The velocity of the object at the end of 2S is 2m / SD. The average velocity of the object in the first 5S is 3m / s


According to the deduction of the law of uniform linear motion △ x = at2, the acceleration of the object is known as a = △ XT2, and the acceleration of the object is known as a = 2 − 112 = 1m / S2 from the displacement of the object in 1s and 2s, so B is correct. Because the displacement of the object in 1s is 1m, according to the displacement time relationship: x = v0t + 12at2, the initial velocity of the object



Grandparents take Xiaoli to the amusement park. It costs 17.5 yuan to buy tickets. It is known that the price of an adult ticket is equal to that of two children tickets


The price of children's tickets
=17.5÷(2×2+1)
=17.5÷5
=3.5 yuan
The price of adult tickets
=3.5×2
=7 yuan



From city a to city B, the original bus needs to run for 4 hours. After the original route is transformed into city expressway, the average speed is increased by 20 km per hour, which can be reached in 3 hours. How many km is the distance between the two cities?


Suppose the distance between the two cities is x km
x/4+20=x/3
x=240
A: the distance between the two cities is 240 kilometers



If one of the following of the equation x & # 178; - 8x + k = 0 is three times of the other, find the value of K


Let a be the root of the equation, then 3a is also the root of the equation
∴a+3a=8
That is, a = 2
Substituting into the equation, we get 4-16 + k = 0
k=12



In addition to substitution elimination method and addition and subtraction elimination method, is there any other method to solve the equations (binary once, ternary once, quaternion once?
I'm from junior high school
Hint: determinant / matrix (this is the key word given by the teacher)
To have a method (teach me), but also have at least two typical examples!
(the better, the more points!)


That is the determinant method, which can get the result directly
The stability of binary linear equations
a1x+b1y=c1
a2x+b2y=c2
Discriminant d = a1b2-a2b1
If d0, there is a unique x = (b2c1-b1c2) / D, y = (a1c2-a2c1) / d
If d = 0, if B 2c1-b 1C 2 = 0, there are innumerable solutions
If B 2c1-b 1C2, then there is no solution
The first order equations of three variables are as follows
a1x+b1y+c1z=d1
a2x+b2y+c2z=d2
a3x+b3y+c3z=d3
There is uniqueness when d 0
x=dx/d, y=dy/d, z=dz/d,
Where D, DX, Dy, DZ are as follows:
|a1 b1 c1 |
d =|a2 b2 c2 |,
|a3 b3 c3 |
|d1 b1 c1 |
dx= |d2 b2 c2 |,
|d3 b3 c3 |
|a1 d1 c1 |
dy =|a2 d2 c2 |,
|a3 d3 c3 |
|a1 b1 d1 |
dz= |a2 b2 d2 |,
|a3 b3 d3 |



In the equal ratio sequence an, A3 = 1 / 2, A9 = 8, then the value of A5 * A6 * A7 is


A9/A3=(A1×q^8)/(A1×q^2)=q^6=8/(1/2)=16
q^3=±4
A5×A6×A7
=A1×q^4×A1×q^5×A1×q^6
=(A1)^3×q^15
=(A1×q^5)^3
=(A1×q^2×q^3)^3
=(A3×q^3)^3
=(1/2×(±4))^3
=±8



8 / 25 square meters = () square decimeters


8 / 25 square meters = (32) square decimeters
8÷25×100=32